Projectile Motion PYQs: MHT CET, JEE Main, NEET

Q1JEE Main 2025 · 24 JAN · SHIFT 2Easy

Motion in a Plane (2-D Kinematics) · Position, velocity & acceleration vectors

The position vector of a moving body at any instant of time is given as $\vec{r} = \left(5t^{2}\hat{i} - 5t\hat{j}\right)\text{ m}$ . The magnitude and direction of velocity at $t = 2\text{ s}$ is,

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Answer: (a)

$\vec v = \frac{d\vec r}{dt} = 10t\,\hat i - 5\,\hat j$ ; at $t = 2$ s, $\vec v = 20\hat i - 5\hat j$
$|\vec v| = \sqrt{400 + 25} = \sqrt{425} = 5\sqrt{17}$ m/s
Angle with $-y$ axis: $\tan\phi = \frac{20}{5} = 4$
So the answer is (a) $5\sqrt{17}$ m/s at $\tan^{-1}4$ with $-$ ve Y axis.

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Q2JEE Main 2024 · 27 JAN · SHIFT 1Easy

Motion in a Plane (2-D Kinematics) · Position, velocity & acceleration vectors

Position of an ant ( $\text{S}$ in metres) moving in $\text{Y-Z}$ plane is given by $S = 2t^{2}\hat{j} + 5\hat{k}$ (where $t$ is in second). The magnitude and direction of velocity of the ant at $t = 1\text{ s}$ will be:

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Answer: (d)

$\vec v = \frac{d\vec S}{dt} = 4t\,\hat j$ (the $5\hat k$ term is constant)
At $t = 1$ s, $\vec v = 4\hat j$ m/s
So the answer is (d) 4 m/s in $y$ -direction.

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Q3JEE Main 2024 · 4 APR · SHIFT 1Medium

Motion in a Plane (2-D Kinematics) · Path from x(t) and y(t)

The co-ordinates of a particle moving in $x\text{-}y$ plane are given by: $x = 2 + 4t,\ y = 3t + 8t^{2}$ . The motion of the particle is:

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Answer: (d)

$v_x = 4$ (constant), $v_y = 3 + 16t$ , so $\vec a = 16\hat j$ (constant): uniformly accelerated
Initial velocity $4\hat i + 3\hat j$ is not along $\vec a$ , so the path is curved
Eliminating $t = \frac{x-2}{4}$ gives $y$ quadratic in $x$ : a parabola
So the answer is (d).

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Q4JEE Main 2024 · 27 JAN · SHIFT 1Medium

Motion in a Plane (2-D Kinematics) · Constant acceleration in a plane

A particle starts from origin at $t = 0$ with a velocity $5\hat{i}\text{ m/s}$ and moves in $x - y$ plane under action of a force which produces a constant acceleration of $(3\hat{i} + 2\hat{j})\text{ m/s}^{2}$ . If the $x$ -coordinate of the particle at that instant is $84\text{ m}$ , then the speed of the particle at this time is $\sqrt{\alpha}\text{ m/s}$ . The value of $\alpha$ is _________.

Numerical answer type. Work it out, then open the answer.

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Answer: 673

$x = 5t + \frac32t^2 = 84 \Rightarrow 3t^2 + 10t - 168 = 0 \Rightarrow t = 6$ s
$v_x = 5 + 3(6) = 23$ m/s, $v_y = 2(6) = 12$ m/s
$v^2 = 529 + 144 = 673$
So $\alpha = 673$ .

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Q5JEE Main 2022 · 28 JUL · SHIFT 2Medium

Motion in a Plane (2-D Kinematics) · Constant acceleration in a plane

At time $t = 0$ a particle starts travelling from a height $7\hat{z}\text{ cm}$ in a plane keeping z coordinate constant. At any instant of time its position along the $\hat{x}$ and $\hat{y}$ directions are defined as $3t$ and $5t^{3}$ respectively. At t = 1s acceleration of the particle will be

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Answer: (b)

$x = 3t \Rightarrow a_x = 0$ ; $z = 7$ constant $\Rightarrow a_z = 0$
$y = 5t^3 \Rightarrow a_y = 30t = 30$ at $t = 1$ s
So the answer is (b) $30\hat y$ .

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Q6JEE Main 2021 · 16 MAR · SHIFT 2Medium

Motion in a Plane (2-D Kinematics) · Constant acceleration in a plane

A mosquito is moving with a velocity $\vec{v} = 0.5t^{2}\hat{i} + 3t\hat{j} + 9\hat{k}$ m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2 s?

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Answer: (a)

At $t = 2$ s: $\vec v = 0.5(4)\hat i + 3(2)\hat j + 9\hat k = 2\hat i + 6\hat j + 9\hat k$
Angle $\phi$ from y-axis: $\tan\phi = \frac{\sqrt{v_x^2 + v_z^2}}{v_y} = \frac{\sqrt{4 + 81}}{6} = \frac{\sqrt{85}}{6}$
So the answer is (a) $\tan^{-1}\left(\frac{\sqrt{85}}{6}\right)$ from y-axis.

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Q7JEE Main 2020 · 8 JAN · SHIFT 2Medium

Motion in a Plane (2-D Kinematics) · Position, velocity & acceleration vectors

A particle moves such that its position vector $\vec{r}(t) = \cos \omega t\, \hat{i} + \sin \omega t\, \hat{j}$ where $\omega$ is a constant and t is time. Then which of the following statements is true for the velocity $\vec{v}(t)$ and acceleration $\vec{a}(t)$ of the particle:

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Answer: (c)

$\vec v = \omega(-\sin\omega t\,\hat i + \cos\omega t\,\hat j)$ , and $\vec v\cdot\vec r = 0$ , so $\vec v \perp \vec r$
$\vec a = -\omega^2\vec r$ , directed towards the origin
So the answer is (c).

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Q8JEE Main 2020 · 8 JAN · SHIFT 1Medium

Motion in a Plane (2-D Kinematics) · Position, velocity & acceleration vectors

A particle is moving along the x-axis with its coordinate with the time 't' given be x(t) = 10 + 8t – 3t2. Another particle is moving the y-axis with its coordinate as a function of time given by y(t) = 5 – 8t3.

At t = 1s, the speed of the second particle as measured in the frame of the first particle is given as $\sqrt{v}$ . Then v (in m/s) is ______.

Numerical answer type. Work it out, then open the answer.

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Answer: 580

First particle: $v_1 = \frac{dx}{dt} = 8 - 6t = 2$ m/s along $x$ at $t = 1$ s
Second particle: $v_2 = \frac{dy}{dt} = -24t^2 = -24$ m/s along $y$ at $t = 1$ s
$\vec v_{21} = \vec v_2 - \vec v_1 = -2\hat i - 24\hat j$
$|\vec v_{21}|^2 = 4 + 576 = 580$ , so $v = 580$ m/s (in the sense $\sqrt v$ m/s).

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Q9JEE Main 2020 · 5 SEP · SHIFT 1Medium

Motion in a Plane (2-D Kinematics) · Path from x(t) and y(t)

A balloon is moving up in air vertically above a point A on the ground. When it is at a height h1, a girl standing at a distance d (point B) from A (see figure) sees it at an angle 45o with respect to the vertical. When the balloon climbs up a further height h2, it is seen at an angle 60o with respect to the vertical if the girl moves further by a distance 2.464 d (point C). Then the height h2 is (given tan 30o = 0.5774)

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Answer: (b)

At B (45 $^\circ$ with vertical): $\tan45^\circ = \frac{d}{h_1} \Rightarrow h_1 = d$
At C (60 $^\circ$ with vertical): $\tan60^\circ = \frac{d + 2.464d}{h_1 + h_2} = \frac{3.464d}{h_1 + h_2}$
$h_1 + h_2 = \frac{3.464d}{1.732} = 2d \Rightarrow h_2 = d$
So the answer is (b) $d$ .

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Q10JEE Main 2020 · 4 SEP · SHIFT 1Medium

Motion in a Plane (2-D Kinematics) · Constant acceleration in a plane

Starting from the origin at time t = 0, with initial velocity 5 $\hat{j}$ ms-1, a particle moves in the x-y plane with a constant acceleration of $\left(10\hat{i} + 4\hat{j}\right)$ ms-2. At time t, its coordinates are (20 m, y0 m). The values of t and y0 are, respectively:

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Answer: (b)

$x = \frac12(10)t^2 = 5t^2 = 20 \Rightarrow t = 2$ s
$y_0 = 5t + \frac12(4)t^2 = 10 + 8 = 18$ m
So the answer is (b) 2 s and 18 m.

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Q11JEE Main 2020 · 9 JAN · SHIFT 2Medium

Motion in a Plane (2-D Kinematics) · Constant acceleration in a plane

A particle starts from the origin at t = 0 with an

initial velocity of 3.0 $\hat{i}$ m/s and moves in the

x-y plane with a constant acceleration $\left(6\hat{i} + 4\hat{j}\right)$ m/s2. The x-coordinate of the particle at the instant when its y-coordinate is 32 m is D meters. The value of D is :-

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Answer: (d)

$y = \frac12(4)t^2 = 2t^2 = 32 \Rightarrow t = 4$ s
$x = 3t + \frac12(6)t^2 = 12 + 48 = 60$ m
So the answer is (d) $D = 60$ .

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Q12JEE Main 2019 · 9 JAN · SHIFT 2Medium

Motion in a Plane (2-D Kinematics) · Position, velocity & acceleration vectors

The position co-ordinates of a particle moving in a 3-D coordinate system is given by x = a cos $\omega$ t

y = a sin $\omega$ t and

z = a $\omega$ t

The speed of the particle is:

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Answer: (a)

$v_x = -a\omega\sin\omega t$ , $v_y = a\omega\cos\omega t$ , $v_z = a\omega$
$v = \sqrt{a^2\omega^2(\sin^2\omega t + \cos^2\omega t) + a^2\omega^2} = \sqrt2\,a\omega$
So the answer is (a) $\sqrt2\,a\omega$ .

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Q13JEE Main 2019 · 9 JAN · SHIFT 1Medium

Motion in a Plane (2-D Kinematics) · Path from x(t) and y(t)

A particle is moving with a velocity

$\vec{v} = K(y\hat{i} + x\hat{j}),$ where K is a constant.

The general equation for its path is:

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Answer: (c)

$\frac{dx}{dt} = Ky$ , $\frac{dy}{dt} = Kx \Rightarrow \frac{dy}{dx} = \frac{x}{y}$
$y\,dy = x\,dx \Rightarrow y^2 = x^2 + \text{constant}$
So the answer is (c).

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Q14NEET 2017Easy

Motion in a Plane (2-D Kinematics) · Constant acceleration in a plane

The x and y coordinates of the particle at any time are x = 5t $-$ 2t2 and y = 10t respectively, where x and y are in metres and t in seconds. The acceleration of the particle at t = 2 s is

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Answer: (b)

$a_x = \frac{d^2x}{dt^2} = -4$ m/s $^2$ , $a_y = \frac{d^2y}{dt^2} = 0$
Acceleration is constant, so at $t = 2$ s it is $-4$ m/s $^2$ .
So the answer is (b) $-4$ m s $^{-2}$ .

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Q15NEET 2016 · PHASE 1Medium

Motion in a Plane (2-D Kinematics) · Position, velocity & acceleration vectors

A particle moves so that its position vector is given by $\vec{r} = \cos \omega t\, \hat{x} + \sin \omega t\, \hat{y},$ where $\omega$ is a constant. Which of the following is true?

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Answer: (a)

$\vec v = -\omega\sin\omega t\,\hat x + \omega\cos\omega t\,\hat y$ , and $\vec v\cdot\vec r = 0$ , so $\vec v \perp \vec r$
$\vec a = \frac{d\vec v}{dt} = -\omega^2\vec r$ , so $\vec a$ points towards the origin
So the answer is (a).

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Q16NEET 2015 · AIPMTMedium

Motion in a Plane (2-D Kinematics) · Position, velocity & acceleration vectors

The position vector of a particle $\vec{R}$ as a function of time is given by $\vec{R}$ = 4sin(2 $\pi t)\hat{i}$ + 4cos(2 $\pi t)\hat{j}$ . Where R is in meters, t is in seconds and $\hat{i}$ and $\hat{j}$ denote unit vectors along x-and y-directions, respectively. Which one of the following statements is wrong for the motion of particle?

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Answer: (a)

$|\vec R| = 4$ m always, so the path is a circle of radius 4 m: (b) true.
$\vec v = 8\pi\cos(2\pi t)\hat i - 8\pi\sin(2\pi t)\hat j$ , so $|\vec v| = 8\pi$ m/s, not 8 m/s: (a) wrong.
$\vec a = -4\pi^2\vec R$ , along $-\vec R$ with magnitude $\frac{v^2}{R}$ : (c), (d) true.
So the wrong statement is (a).

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Q17NEET 2014 · AIPMTEasy

Motion in a Plane (2-D Kinematics) · Average velocity & acceleration

A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m) at time t = 2 s and (13 m, 14 m) at time t = 5 s.

Average velocity vector $\left(\vec{v}_{av}\right)$ from t = 0 to t = 5 s is

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Answer: (d)

$\vec v_{av} = \frac{\Delta\vec r}{\Delta t} = \frac{(13-2)\hat i + (14-3)\hat j}{5 - 0}$
$= \frac{11}{5}(\hat i + \hat j)$
So the answer is (d) $\frac{11}{5}(\hat i + \hat j)$ .

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Q18NEET 2012 · AIPMT · PRELIMSEasy

Motion in a Plane (2-D Kinematics) · Constant acceleration in a plane

A particle has initial velocity $\left(2\vec{i} + 3\vec{j}\right)$ and acceleration $\left(0.3\vec{i} + 0.2\vec{j}\right)$ . The magnitude of velocity after 10 seconds will be

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Answer: (b)

$\vec v = \vec u + \vec a t = (2 + 0.3\times10)\hat i + (3 + 0.2\times10)\hat j = 5\hat i + 5\hat j$
$|\vec v| = \sqrt{25 + 25} = 5\sqrt2$
So the answer is (b) $5\sqrt2$ .

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Q19NEET 2011 · AIPMT · PRELIMSEasy

Motion in a Plane (2-D Kinematics) · Average velocity & acceleration

A body is moving with velocity 30 m/s towards east. After 10 seconds its velocity becomes 40 m/s towards north. The average acceleration of the body is

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Answer: (d)

$\Delta\vec v = 40\,\hat N - 30\,\hat E$ , so $|\Delta\vec v| = \sqrt{40^2 + 30^2} = 50$ m/s
$a_{av} = \frac{|\Delta\vec v|}{t} = \frac{50}{10} = 5$ m/s $^2$
So the answer is (d) 5 m/s $^2$ .

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Q20NEET 2010 · AIPMT · MAINSEasy

Motion in a Plane (2-D Kinematics) · Path from x(t) and y(t)

A particle moves in x-y plane according to rule x = asin $\omega$ t and y = acos $\omega$ t. The particle follows

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Answer: (b)

$x^2 + y^2 = a^2\sin^2\omega t + a^2\cos^2\omega t = a^2$
This is a circle of radius $a$ centred at the origin.
So the answer is (b) a circular path.

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Q21NEET 2010 · AIPMT · PRELIMSEasy

Motion in a Plane (2-D Kinematics) · Constant acceleration in a plane

A particle has initial velocity $\left(3\hat{i} + 4\hat{j}\right)$ and has acceleration $\left(0.4\hat{i} + 0.3\hat{j}\right).$ Its speed after 10 s is

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Answer: (b)

$\vec v = \vec u + \vec a t = (3 + 4)\hat i + (4 + 3)\hat j = 7\hat i + 7\hat j$
Speed $= \sqrt{49 + 49} = 7\sqrt2$ units
So the answer is (b) $7\sqrt2$ units.

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Q22JEE Main 2010 · AIEEEMedium

Motion in a Plane (2-D Kinematics) · Path from x(t) and y(t)

A particle is moving with velocity $\vec{v} = k\left(y\hat{i} + x\hat{j}\right)$ , where K is a constant. The general equation for its path is

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Answer: (d)

$\frac{dx}{dt} = ky$ , $\frac{dy}{dt} = kx \Rightarrow \frac{dy}{dx} = \frac{x}{y}$
$y\,dy = x\,dx \Rightarrow y^2 = x^2 + \text{constant}$
So the answer is (d).

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Q23JEE Main 2009 · AIEEEEasy

Motion in a Plane (2-D Kinematics) · Constant acceleration in a plane

A particle has an initial velocity $3\hat{i} + 4\hat{j}$ and an acceleration of $0.4\hat{i} + 0.3\hat{j}$ . Its speed after 10 s is:

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Answer: (a)

$\vec v = \vec u + \vec a t = (3 + 0.4\times10)\hat i + (4 + 0.3\times10)\hat j = 7\hat i + 7\hat j$
Speed $= \sqrt{49 + 49} = 7\sqrt2$ units
So the answer is (a) $7\sqrt2$ units.

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Q24NEET 2007 · AIPMTEasy

Motion in a Plane (2-D Kinematics) · Path from x(t) and y(t)

A particle starting from the origin (0, 0) moves in a straight line in the (x, y) plane. Its coordinates at a later time are $\left(\sqrt{3}, 3\right)$ . The path of the particle makes with the x-axis an angle of

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Answer: (b)

Slope of the straight path: $\tan\theta = \frac{y}{x} = \frac{3}{\sqrt3} = \sqrt3$
So the answer is (b) $60^\circ$ .

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Q25JEE Main 2005 · AIEEEEasy

Motion in a Plane (2-D Kinematics) · Average velocity & acceleration

A particle is moving eastwards with a velocity of 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is

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Answer: (c)

$\Delta\vec v = \vec v_2 - \vec v_1 = 5\,\hat N - 5\,\hat E = 5\,\hat N + 5\,\hat W$
$|\Delta\vec v| = 5\sqrt2$ m/s, directed north-west
$a_{av} = \frac{5\sqrt2}{10} = \frac{1}{\sqrt2}$ m s $^{-2}$
So the answer is (c) $\frac{1}{\sqrt2}$ m s $^{-2}$ towards north-west.

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Q26NEET 1998 · AIPMTMedium

Motion in a Plane (2-D Kinematics) · Path from x(t) and y(t)

Two particles $A$ and $B$ are connected by a rigid rod $AB$ . The rod slides along perpendicular rails as shown here. The velocity of $A$ to the right is $10\text{ m/s}$ . What is the velocity of $B$ when angle $\alpha = 60^\circ$ ?

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Answer: (c)

Rod length fixed: $x^2 + y^2 = L^2 \Rightarrow x\frac{dx}{dt} + y\frac{dy}{dt} = 0$
$|v_B| = v_A\frac{x}{y} = v_A\cot\alpha$
$|v_B| = 10\cot60^\circ = \frac{10}{\sqrt3} \approx 5.8$ m/s (downward)
So the answer is (c) 5.8 m/s.

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Q27NEET 1995 · AIPMTEasy

Motion in a Plane (2-D Kinematics) · Position, velocity & acceleration vectors

The position vector of a particle is $\vec{r} = (a \cos \omega t)\hat{i} + (a \sin \omega t)\hat{j}$ . The velocity of the particle is

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Answer: (d)

$\vec v = \frac{d\vec r}{dt} = -a\omega\sin\omega t\,\hat i + a\omega\cos\omega t\,\hat j$
$\vec v\cdot\vec r = -a^2\omega\sin\omega t\cos\omega t + a^2\omega\sin\omega t\cos\omega t = 0$
So the answer is (d): velocity is perpendicular to the position vector.

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Q28NEET 1989 · AIPMTMedium

Motion in a Plane (2-D Kinematics) · Average velocity & acceleration

A bus is moving on a straight road towards north with a uniform speed of 50 km/hour then it turns left through $90^\circ$ . If the speed remains unchanged after turning, the increase in the velocity of bus in the turning process is

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Answer: (a)

$\Delta\vec v = \vec v_{west} - \vec v_{north} = 50\,\hat W + 50\,\hat S$
$|\Delta\vec v| = 50\sqrt2 \approx 70.7$ km/h, directed south-west
So the answer is (a) 70.7 km/hr along south-west.

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Q29MHT CET 2026 · 19 APR · SHIFT 2Medium

Projectile Basics & Trajectory · Equation of trajectory

The equation of the trajectory of a ball projected at an angle $\theta$ with the horizontal, is given as $y = x - \frac{gx^{2}}{2}$

The initial velocity of the ball is

[Given: $\tan 45^\circ = 1,\ \cos 45^\circ = \frac{1}{\sqrt{2}}$ ]

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Answer: (c)

Compare with $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$ : $\tan\theta = 1 \Rightarrow \theta = 45^\circ$
$\frac{g}{2u^2\cos^2\theta} = \frac g2 \Rightarrow u^2\cos^2\theta = 1 \Rightarrow u^2 \cdot \frac12 = 1$
$u = \sqrt2$ m/s
So the answer is (c) $\sqrt2$ m/s.

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Q30JEE Main 2023 · 8 APR · SHIFT 2Medium

Projectile Basics & Trajectory · Equation of trajectory

The trajectory of projectile, projected from the ground is given by $y = x - \frac{x^{2}}{20}$ . Where $x$ and $y$ are measured in meter. The maximum height attained by the projectile will be.

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Answer: (b)

$y = 0$ at $x = 0$ and $x = 20$ , so range $R = 20$ m; maximum at $x = 10$ m
$H = 10 - \frac{100}{20} = 5$ m
So the answer is (b) 5 m.

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Q31JEE Main 2022 · 26 JUL · SHIFT 1Medium

Projectile Basics & Trajectory · Equation of trajectory

If the initial velocity in horizontal direction of a projectile is unit vector $\hat{i}$ and the equation of trajectory $y = 5x(1 - x)$ . The $y$ component vector of the initial velocity is ______________ $\hat{j}$ . ( $\text{Take } g = 10\text{ m/s}^{2})$

Numerical answer type. Work it out, then open the answer.

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Answer: 5

$y = 5x - 5x^2$ ; compare with $y = x\frac{u_y}{u_x} - \frac{gx^2}{2u_x^2}$
$\frac{g}{2u_x^2} = \frac{10}{2(1)^2} = 5$ (consistent with $u_x = 1$ )
$\frac{u_y}{u_x} = 5 \Rightarrow u_y = 5$
So the $y$ component is $5\hat j$ m/s, answer 5.

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Q32JEE Main 2021 · 26 FEB · SHIFT 2Medium

Projectile Basics & Trajectory · Equation of trajectory

The trajectory of a projectile in a vertical plane is y = $\alpha$ x $-\beta$ x2, where $\alpha$ and $\beta$ are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $\theta$ and the maximum height attained H are respectively given by:

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Answer: (a)

Compare with $y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$ : $\tan\theta = \alpha \Rightarrow \theta = \tan^{-1}\alpha$
$\frac{dy}{dx} = 0$ at $x = \frac{\alpha}{2\beta}$
$H = \alpha\cdot\frac{\alpha}{2\beta} - \beta\frac{\alpha^2}{4\beta^2} = \frac{\alpha^2}{4\beta}$
So the answer is (a).

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Q33JEE Main 2019 · 12 APR · SHIFT 1Medium

Projectile Basics & Trajectory · Equation of trajectory

The trajectory of a projectile near the surface of the earth is given as y = 2x – 9x2. If it were launched at an angle $\theta$ 0 with speed v0 then (g = 10 ms–2):

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Answer: (a)

$\tan\theta_0 = 2 \Rightarrow \cos\theta_0 = \frac{1}{\sqrt5}$
$\frac{g}{2v_0^2\cos^2\theta_0} = 9 \Rightarrow \frac{10}{2v_0^2/5} = 9 \Rightarrow v_0^2 = \frac{25}{9}$
$v_0 = \frac53$ m/s
So the answer is (a).

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Q34JEE Main 2014 · 9 APR · SHIFT 1Medium

Projectile Basics & Trajectory · Position at any time

The position of a projectile launched from the origin at t = 0 is given by $\vec{r}$ = (40 $\hat{i}$ + 50 $\hat{j}$ ) m at t = 2s. If the projectile was launched at an angle $\theta$ from the horizontal, then $\theta$ is (take g = 10 ms ${}^{-2}$ ).

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Answer: (c)

$x = u_x t \Rightarrow 40 = 2u_x \Rightarrow u_x = 20$ m/s
$y = u_y t - \frac12gt^2 \Rightarrow 50 = 2u_y - 20 \Rightarrow u_y = 35$ m/s
$\tan\theta = \frac{u_y}{u_x} = \frac{35}{20} = \frac74$
So the answer is (c) $\tan^{-1}\frac74$ .

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Q35JEE Main 2013Easy

Projectile Basics & Trajectory · Equation of trajectory

A projectile is given an initial velocity of $\left(\hat{i} + 2\hat{j}\right)$ m/s, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. If g = 10 m/s2, the equation of its trajectory is:

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Answer: (b)

$u_x = 1$ , $u_y = 2$ : $x = t$ , $y = 2t - 5t^2$
Put $t = x$ : $y = 2x - 5x^2$
So the answer is (b).

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Q36JEE Main 2013 · 22 APR · SHIFT 1Difficult

Projectile Basics & Trajectory · Equation of trajectory

A ball projected from ground at an angle of 45° just clears a wall in front. If point of projection is 4 m from the foot of wall and ball strikes the ground at a distance of 6 m on the other side of the wall, the height of the wall is:

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Answer: (b)

Range $R = 4 + 6 = 10$ m; trajectory $y = x\tan\theta\left(1 - \frac xR\right)$
At the wall $x = 4$ m, $\theta = 45^\circ$ : $y = 4\times1\times\left(1 - \frac{4}{10}\right)$
$y = 2.4$ m
So the answer is (b) 2.4 m.

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Q37NEET 2011 · AIPMT · MAINSMedium

Projectile Basics & Trajectory · Elevation of the highest point

A projectile is fired at an angle of 45o with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is

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Answer: (c)

Highest point is at $\left(\frac R2, H\right)$ , so $\tan\alpha = \frac{H}{R/2} = \frac{2H}{R}$
$\frac HR = \frac{\tan\theta}{4} \Rightarrow \tan\alpha = \frac{\tan\theta}{2} = \frac12$
So the answer is (c) $\tan^{-1}\left(\frac12\right)$ .

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Q38JEE Main 2003 · AIEEEMedium

Projectile Basics & Trajectory · Position at any time

A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of $30^\circ$ with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? $\left[g = 10\text{ m/s}^{2},\ \sin 30^\circ = \frac{1}{2},\ \cos 30^\circ = \frac{\sqrt{3}}{2}\right]$

Show answer and solution

Answer: (d)

Height 10 m from ground is the level of the throwing point, so we need the horizontal range on that level.
$R = \frac{u^2\sin2\theta}{g} = \frac{100\sin60^\circ}{10} = 10\times\frac{\sqrt3}{2}$
$R = 8.66$ m
So the answer is (d) 8.66 m.

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Q39MHT CET 2026 · 15 APR · SHIFT 2Medium

Velocity, Momentum & Energy in Flight · Velocity at any instant

A batsman hits a ball with a velocity 'v', making an angle of $60^\circ$ with the vertical. After some time direction of velocity is making an angle of $60^\circ$ with the horizontal. The speed of the ball at this instant is

$\left[\cos(60^\circ) = \frac{1}{2},\ \cos(30^\circ) = \frac{\sqrt{3}}{2}\right]$

Show answer and solution

Answer: (b)

$60^\circ$ with the vertical means $30^\circ$ with the horizontal, so $v_x = v\cos 30^\circ = \frac{\sqrt{3}}{2}v$ .
The horizontal component stays constant throughout the flight.
When the velocity makes $60^\circ$ with the horizontal: $v'\cos 60^\circ = \frac{\sqrt{3}}{2}v$
$v' = \frac{\sqrt{3}v/2}{1/2} = \sqrt{3}v$
So the answer is (b) $\sqrt{3}v$ .

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Q40JEE Main 2026 · 4 APR · SHIFT 2Medium

Velocity, Momentum & Energy in Flight · Velocity at any instant

If $x$ and $y$ coordinates of a projectile as a function of time $(t)$ are given as $24t$ and $43.6t - 4.9t^{2}$ , respectively, then the angle (in degrees) made by the projectile with horizontal when $t = 2\text{ s}$ is ______.

Show answer and solution

Answer: (b)

$v_x = \frac{dx}{dt} = 24\text{ m/s}$ , $v_y = \frac{dy}{dt} = 43.6 - 9.8t$
At $t = 2\text{ s}$ : $v_y = 43.6 - 19.6 = 24\text{ m/s}$
$\tan\alpha = \frac{v_y}{v_x} = \frac{24}{24} = 1 \Rightarrow \alpha = 45^\circ$
So the answer is (b) 45.

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Q41JEE Main 2026 · 22 JAN · SHIFT 1Medium

Velocity, Momentum & Energy in Flight · Velocity at any instant

A projectile is thrown upward at an angle $60^\circ$ with the horizontal. The speed of the projectile is $20\text{ m/s}$ when its direction of motion is $45^\circ$ with the horizontal. The initial speed of the projectile is $\underline{\qquad}\text{ m/s}$ .

Show answer and solution

Answer: (b)

Horizontal component is constant: $u\cos 60^\circ = v\cos 45^\circ$
$u \times \frac{1}{2} = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2}$
$u = 20\sqrt{2}\text{ m/s}$
So the answer is (b) $20\sqrt{2}$ .

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Q42MHT CET 2026 · 13 MAY · SHIFT 1Medium

Velocity, Momentum & Energy in Flight · Velocity at the highest point

A particle is projected from ground with speed u and at an angle $\theta$ with horizontal. If at maximum height from ground, the speed of a particle is half times its initial velocity of projection then its maximum height attained is $(\cos 60^\circ = 0.5,\ \sin 60^\circ = \sqrt{3}/2)$ (g = acceleration due to gravity)

Show answer and solution

Answer: (b)

Speed at the top $= u\cos\theta = \frac{u}{2} \Rightarrow \theta = 60^\circ$
$H = \frac{u^2\sin^2\theta}{2g} = \frac{u^2 (3/4)}{2g}$
$H = \frac{3u^2}{8g}$
So the answer is (b).

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Q43MHT CET 2026 · 17 APR · SHIFT 2Medium

Velocity, Momentum & Energy in Flight · Energy in flight

A ball P is projected at an angle of $60^\circ$ with the vertical with certain initial speed. Another ball Q of the same mass as that of ball P is projected vertically upwards with the same initial speed as that of P. At the highest point, the ratio of potential energy of ball P to that of ball Q is

$(\sin 30^\circ = 0.5)$

Show answer and solution

Answer: (a)

P makes $30^\circ$ with the horizontal: $H_P = \frac{u^2\sin^2 30^\circ}{2g} = \frac{u^2}{8g}$
Q goes straight up: $H_Q = \frac{u^2}{2g}$
PE $\propto$ height (same mass): $\frac{U_P}{U_Q} = \frac{1/8}{1/2} = \frac{1}{4}$
So the answer is (a) $1:4$ .

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Q44JEE Main 2025 · 22 JAN · SHIFT 2Easy

Velocity, Momentum & Energy in Flight · Energy in flight

A ball of mass 100 g is projected with velocity $20\text{ m/s}$ at $60^\circ$ with horizontal. The decrease in kinetic energy of the ball during the motion from point of projection to highest point is

Show answer and solution

Answer: (c)

Loss of KE up to the top equals the KE of the vertical component: $\Delta K = \frac{1}{2}m(u\sin\theta)^2$ .
$u\sin 60^\circ = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\text{ m/s}$
$\Delta K = \frac{1}{2} \times 0.1 \times 300 = 15\text{ J}$
So the answer is (c) 15 J.

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Q45JEE Main 2025 · 24 JAN · SHIFT 1Medium

Velocity, Momentum & Energy in Flight · Angular momentum

An object of mass ' m ' is projected from origin in a vertical xy plane at an angle $45^\circ$ with the $x\text{-}$ axis with an initial velocity $v_0$ . The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches at the maximum height, will be [ g is acceleration due to gravity]

Show answer and solution

Answer: (d)

At the top, $\vec r = x\hat i + H\hat j$ and $\vec p = m v_0\cos 45^\circ\,\hat i$ , so $|\vec L| = m(v_0\cos 45^\circ)H$ .
$H = \frac{v_0^2\sin^2 45^\circ}{2g} = \frac{v_0^2}{4g}$
$L = m \cdot \frac{v_0}{\sqrt{2}} \cdot \frac{v_0^2}{4g} = \frac{m v_0^3}{4\sqrt{2}g}$
$\vec L = (H\hat j) \times (p\,\hat i) = -Hp\,\hat k$ , i.e. along negative $z$ -axis.
So the answer is (d).

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Q46JEE Main 2023 · 1 FEB · SHIFT 2Easy

Velocity, Momentum & Energy in Flight · Velocity at any instant

For a body projected at an angle with the horizontal from the ground, choose the correct statement.

Show answer and solution

Answer: (a)

(a) Height is greatest at the top, so gravitational PE is maximum there: true.
(b) Vertical velocity and momentum are zero at the top: false.
(c) Horizontal velocity stays $u\cos\theta \ne 0$ : false. (d) KE $= \frac12m u^2\cos^2\theta \ne 0$ at the top: false.
So the answer is (a).

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Q47NEET 2023Easy

Velocity, Momentum & Energy in Flight · Velocity at the highest point

A ball is projected from point A with velocity $20\text{ m s}^{-1}$ at an angle $60^\circ$ to the horizontal direction. At the highest point $\text{B}$ of the path (as shown in figure), the velocity $v\text{ ms}^{-1}$ of the ball will be:

Show answer and solution

Answer: (d)

At the highest point the vertical velocity is zero; only the horizontal component $u\cos\theta$ remains.
$v = 20\cos 60^\circ = 20 \times \frac{1}{2} = 10\text{ m s}^{-1}$
So the answer is (d) 10.

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Q48JEE Main 2023 · 11 APR · SHIFT 2Medium

Velocity, Momentum & Energy in Flight · Velocity at any instant

A projectile is projected at $30^\circ$ from horizontal with initial velocity $40\text{ ms}^{-1}$ . The velocity of the projectile at $t = 2\text{ s}$ from the start will be: (Given $g = 10\text{ m/s}^{2}$ )

Show answer and solution

Answer: (a)

$v_x = 40\cos 30^\circ = 20\sqrt{3}\text{ m/s}$ (constant)
$v_y = 40\sin 30^\circ - gt = 20 - 10 \times 2 = 0$
So $v = 20\sqrt{3}\text{ m/s}$ (the projectile is at the top), answer (a).

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Q49NEET 2022 · PHASE 1Easy

Velocity, Momentum & Energy in Flight · Velocity at the highest point

A ball is projected with a velocity, 10 ms ${}^{-1}$ , at an angle of 60 $^\circ$ with the vertical direction. Its speed at the highest point of its trajectory will be

Show answer and solution

Answer: (b)

$60^\circ$ with the vertical means $\theta = 30^\circ$ with the horizontal.
At the top, speed $= u\cos\theta = 10\cos 30^\circ = 10 \times \frac{\sqrt{3}}{2}$
So the answer is (b) $5\sqrt{3}\text{ m s}^{-1}$ .

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Q50JEE Main 2022 · 27 JUN · SHIFT 1Medium

Velocity, Momentum & Energy in Flight · Velocity at any instant

A projectile is launched at an angle ' $\alpha$ ' with the horizontal with a velocity 20 ms ${}^{-1}$ . After 10 s, its inclination with horizontal is ' $\beta$ '. The value of tan $\beta$ will be: (g = 10 ms ${}^{-2}$ ).

Show answer and solution

Answer: (b)

$v_x = 20\cos\alpha$ , $v_y = 20\sin\alpha - 10 \times 10 = 20\sin\alpha - 100$
$\tan\beta = \frac{v_y}{v_x} = \frac{20\sin\alpha - 100}{20\cos\alpha}$
$\tan\beta = \tan\alpha - 5\sec\alpha$
So the answer is (b).

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Q51JEE Main 2022 · 24 JUN · SHIFT 2Medium

Velocity, Momentum & Energy in Flight · Velocity at the highest point

A body is projected from the ground at an angle of 45 $^\circ$ with the horizontal. Its velocity after 2s is 20 ms ${}^{-1}$ . The maximum height reached by the body during its motion is __________ m. (use g = 10 ms ${}^{-2}$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 20

At $45^\circ$ : $u_x = u_y = \frac{u}{\sqrt{2}}$ . After 2 s: $v_y = \frac{u}{\sqrt{2}} - 20$ .
$v^2 = \frac{u^2}{2} + \left(\frac{u}{\sqrt{2}} - 20\right)^2 = 400$
$u^2 - 20\sqrt{2}\,u = 0 \Rightarrow u = 20\sqrt{2}\text{ m/s}$
$H = \frac{u^2\sin^2 45^\circ}{2g} = \frac{800 \times \frac{1}{2}}{20} = 20\text{ m}$
So the maximum height is 20 m.

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Q52JEE Main 2022 · 27 JUL · SHIFT 2Difficult

Velocity, Momentum & Energy in Flight · Change in velocity & momentum

A body of mass $10\text{ kg}$ is projected at an angle of $45^\circ$ with the horizontal. The trajectory of the body is observed to pass through a point $(20, 10)$ . If $T$ is the time of flight, then its momentum vector, at time $t = \frac{T}{\sqrt{2}}$ , is _____________.

[Take $g = 10\text{ m/s}^{2}$ ]

Show answer and solution

Answer: (d)

Trajectory at $45^\circ$ : $y = x - \frac{g x^2}{u^2}$ . Through $(20, 10)$ : $10 = 20 - \frac{10 \times 400}{u^2} \Rightarrow u = 20\text{ m/s}$
$u_x = u_y = 10\sqrt{2}\text{ m/s}$ , $T = \frac{2u_y}{g} = 2\sqrt{2}\text{ s}$ , so $t = \frac{T}{\sqrt{2}} = 2\text{ s}$
$v_y = 10\sqrt{2} - 20$ , $v_x = 10\sqrt{2}$
$\vec p = m\vec v = 100\sqrt{2}\,\hat i + (100\sqrt{2} - 200)\hat j$
So the answer is (d).

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Q53NEET 2013Easy

Velocity, Momentum & Energy in Flight · Velocity at any instant

The velocity of a projectile at the initial point A is $\left(2\hat{i} + 3\hat{j}\right)$ m/s. Its velocity (in m/s) at point B is

Show answer and solution

Answer: (a)

Horizontal velocity stays $2\hat i$ throughout.
B is at the same level as A, so the vertical velocity has the same size but is reversed: $-3\hat j$
So the answer is (a) $2\hat i - 3\hat j$ .

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Q54NEET 2010 · AIPMT · MAINSEasy

Velocity, Momentum & Energy in Flight · Velocity at the highest point

The speed of a projectile at its maximum height is half of its initial speed. The angle of projection is

Show answer and solution

Answer: (a)

Speed at maximum height $= u\cos\theta$ .
$u\cos\theta = \frac{u}{2} \Rightarrow \cos\theta = \frac{1}{2}$
So the answer is (a) $60^\circ$ .

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Q55MHT CET 2026 · 14 MAY · SHIFT 2Easy

Time of Flight, Maximum Height & Range · Maximum height

If the initial velocity of projection be doubled, keeping the angle of projection same, the maximum height reached by it will

Show answer and solution

Answer: (c)

$H = \frac{u^2\sin^2\theta}{2g} \Rightarrow H \propto u^2$ for the same angle.
Doubling $u$ gives $H' = 2^2 H = 4H$
So the answer is (c) become four times.

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Q56MHT CET 2026 · 18 APR · SHIFT 2Easy

Time of Flight, Maximum Height & Range · Relation between R and H

For a projectile motion, the range R is 'n' times the maximum height H. So the angle of projection is

Show answer and solution

Answer: (c)

$\frac{R}{H} = \frac{2u^2\sin\theta\cos\theta/g}{u^2\sin^2\theta/2g} = \frac{4}{\tan\theta}$
$\frac{4}{\tan\theta} = n \Rightarrow \tan\theta = \frac{4}{n}$
So the answer is (c) $\tan^{-1}\left(\frac{4}{n}\right)$ .

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Q57JEE Main 2026 · 4 APR · SHIFT 2Easy

Time of Flight, Maximum Height & Range · Area covered by bullets

A gun mounted on the ground fires bullets in all directions with same speed. The farthest distance the bullets could reach is 6.4 m. The speed of the bullets from the gun is $\underline{\qquad}\text{ m/s}$ . (take $g = 10\text{ m/s}^{2}$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 8

Farthest reach on level ground is the maximum range: $R_{max} = \frac{u^2}{g}$
$u^2 = 6.4 \times 10 = 64$
$u = 8\text{ m/s}$

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Q58MHT CET 2026 · 14 MAY · SHIFT 1Medium

Time of Flight, Maximum Height & Range · Time of flight

A stone is thrown at an angle $\theta$ with the horizontal reaches a maximum height $H$ , then the time of flight of stone will be ( $g$ = acceleration due to gravity)

Show answer and solution

Answer: (c)

$H = \frac{u^2\sin^2\theta}{2g} \Rightarrow u\sin\theta = \sqrt{2gH}$
$T = \frac{2u\sin\theta}{g} = \frac{2\sqrt{2gH}}{g} = 2\sqrt{\frac{2H}{g}}$
So the answer is (c) $2\left(\frac{2H}{g}\right)^{1/2}$ .

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Q59JEE Main 2026 · 24 JAN · SHIFT 1Medium

Time of Flight, Maximum Height & Range · Maximum height

A boy throws a ball into air at $45^\circ$ from the horizontal to land it on a roof of a building of height $H$ . If the ball attains maximum height in 2 s and lands on the building in 3 s after launch, then value of $H$ is ______ m.

$\left(g = 10\text{ m/s}^{2}\right)$

Show answer and solution

Answer: (d)

Time to reach maximum height $t = \frac{u\sin\theta}{g} = 2\text{ s}$ , so $u\sin 45^\circ = 20\text{ m/s}$ .
Height after $3\text{ s}$ : $y = (u\sin\theta)t - \frac{1}{2}gt^2 = 20(3) - 5(9)$
$H = 60 - 45 = 15\text{ m}$
So the answer is (d) 15.

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Q60MHT CET 2025 · 23 APR · SHIFT 2Medium

Time of Flight, Maximum Height & Range · Time of flight

A body when projected at an angle ' $\theta$ ' with the horizontal reaches a maximum height ' $H$ '. The time of flight of the body will be ( $g =$ acceleration due to gravity)

Show answer and solution

Answer: (c)

$H = \frac{(u\sin\theta)^2}{2g} \Rightarrow u\sin\theta = \sqrt{2gH}$
$T = \frac{2u\sin\theta}{g} = 2\sqrt{\frac{2H}{g}}$
So the answer is (c).

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Q61JEE Main 2025 · 3 APR · SHIFT 1Medium

Time of Flight, Maximum Height & Range · Maximum height

The angle of projection of a particle is measured from the vertical axis as $\phi$ and the maximum height reached by the particle is $h_m$ . Here $h_m$ as function of $\phi$ can be presented as

Show answer and solution

Answer: (d)

Angle from the vertical is $\phi$ , so the vertical component of velocity is $u\cos\phi$ .
$h_m = \frac{u^2\cos^2\phi}{2g}$
At $\phi = 0$ (straight up) $h_m$ is maximum, $\frac{u^2}{2g}$ , and it falls smoothly to zero at $\phi = 90^\circ$ .
The curve starts at its peak with zero slope and drops to zero at $90^\circ$ : graph (d).

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Q62JEE Main 2025 · 3 APR · SHIFT 2Medium

Time of Flight, Maximum Height & Range · Relation between R and H

A particle is projected with velocity $u$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as $\frac{nu^2}{25g}$ , where value of $n$ is: (Given, ' $g$ ' is the acceleration due to gravity.)

Show answer and solution

Answer: (d)

$\frac{R}{H} = \frac{4}{\tan\theta} = 3 \Rightarrow \tan\theta = \frac{4}{3}$
So $\sin\theta = \frac{4}{5}$ , $\cos\theta = \frac{3}{5}$ , $\sin 2\theta = 2\cdot\frac{4}{5}\cdot\frac{3}{5} = \frac{24}{25}$
$R = \frac{u^2\sin 2\theta}{g} = \frac{24u^2}{25g}$
So $n = 24$ , answer (d).

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Q63JEE Main 2025 · 22 JAN · SHIFT 1Difficult

Time of Flight, Maximum Height & Range · Maximum height

A particle is projected at an angle of $30^\circ$ from horizontal at a speed of $60\text{ m/s}$ . The height traversed by the particle in the first second is $h_0$ and height traversed in the last second, before it reaches the maximum height, is $h_1$ . The ratio $h_0 : h_1$ is __________.

[Take, $g = 10\text{ m/s}^{2}$ ]

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 5

Vertical component $u_y = 60\sin 30^\circ = 30\text{ m/s}$ ; time to top $= \frac{30}{10} = 3\text{ s}$ .
First second: $h_0 = 30(1) - \frac{1}{2}(10)(1)^2 = 25\text{ m}$
Last second before the top (reverse motion from rest): $h_1 = \frac{1}{2}(10)(1)^2 = 5\text{ m}$
$h_0 : h_1 = 25 : 5 = 5$

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Q64JEE Main 2024 · 5 APR · SHIFT 2Easy

Time of Flight, Maximum Height & Range · Maximum height

The maximum height reached by a projectile is $64\text{ m}$ . If the initial velocity is halved, the new maximum height of the projectile is ______ $\text{m}$ .

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 16

$H \propto u^2$ for the same angle.
Halving $u$ : $H' = \frac{H}{4} = \frac{64}{4}$
New maximum height $= 16\text{ m}$ .

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Q65MHT CET 2024 · 4 MAY · SHIFT 1Easy

Time of Flight, Maximum Height & Range · Relation between R and H

For a projectile, the maximum height and horizontal range are same. The angle of projection ' $\theta$ ' of the projectile is

Show answer and solution

Answer: (d)

$R = 4H\cot\theta$ (from $\frac{R}{H} = \frac{4}{\tan\theta}$ ).
$R = H \Rightarrow \tan\theta = 4$
So the answer is (d) $\tan^{-1}(4)$ .

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Q66JEE Main 2024 · 8 APR · SHIFT 2Easy

Time of Flight, Maximum Height & Range · Relation between R and H

The angle of projection for a projectile to have same horizontal range and maximum height is:

Show answer and solution

Answer: (d)

$\frac{R}{H} = \frac{4}{\tan\theta}$
$R = H \Rightarrow \tan\theta = 4$
So the answer is (d) $\tan^{-1}(4)$ .

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Q67NEET 2023Easy

Time of Flight, Maximum Height & Range · Maximum height

A bullet is fired from a gun at the speed of $280\text{ ms}^{-1}$ in the direction $30^\circ$ above the horizontal. The maximum height attained by the bullet is

$\left(g = 9.8\text{ ms}^{-2},\ \sin 30^\circ = 0.5\right)$ :-

Show answer and solution

Answer: (b)

$H = \frac{u^2\sin^2\theta}{2g}$
$H = \frac{280^2 \times 0.25}{2 \times 9.8} = \frac{19600}{19.6} = 1000\text{ m}$
So the answer is (b) 1000 m.

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Q68JEE Main 2023 · 24 JAN · SHIFT 1Easy

Time of Flight, Maximum Height & Range · Range & maximum range

The maximum vertical height to which a man can throw a ball is 136 m. The maximum horizontal distance upto which he can throw the same ball is:

Show answer and solution

Answer: (b)

Maximum vertical height (thrown straight up): $\frac{u^2}{2g} = 136\text{ m}$
Maximum range: $R_{max} = \frac{u^2}{g} = 2 \times 136 = 272\text{ m}$
So the answer is (b) 272 m.

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Q69JEE Main 2023 · 6 APR · SHIFT 1Easy

Time of Flight, Maximum Height & Range · Relation between R and H

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion A: When a body is projected at an angle $45^\circ$ , it's range is maximum.

Reason R: For maximum range, the value of $\sin 2\theta$ should be equal to one.

In the light of the above statements, choose the correct answer from the options given below:

Show answer and solution

Answer: (a)

$R = \frac{u^2\sin 2\theta}{g}$ is largest when $\sin 2\theta = 1$ (R is true).
$\sin 2\theta = 1 \Rightarrow 2\theta = 90^\circ \Rightarrow \theta = 45^\circ$ (A is true, and R explains it).
So the answer is (a).

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Q70JEE Main 2023 · 31 JAN · SHIFT 1Medium

Time of Flight, Maximum Height & Range · Time of flight

The initial speed of a projectile fired from ground is $u$ . At the highest point during its motion, the speed of projectile is $\frac{\sqrt{3}}{2}u$ . The time of flight of the projectile is:

Show answer and solution

Answer: (a)

Speed at the top $= u\cos\theta = \frac{\sqrt{3}}{2}u \Rightarrow \theta = 30^\circ$
$T = \frac{2u\sin 30^\circ}{g} = \frac{u}{g}$
So the answer is (a) $\frac{u}{g}$ .

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Q71JEE Main 2023 · 11 APR · SHIFT 1Medium

Time of Flight, Maximum Height & Range · Same height at two instants

A projectile fired at $30^\circ$ to the ground is observed to be at same height at time $3\text{ s}$ and $5\text{ s}$ after projection, during its flight. The speed of projection of the projectile is ___________ $\text{m s}^{-1}$ . (Given $g = 10\text{ m s}^{-2}$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 80

Same height at $t_1$ and $t_2$ means time of flight $T = t_1 + t_2 = 3 + 5 = 8\text{ s}$ .
$T = \frac{2u\sin\theta}{g} \Rightarrow 8 = \frac{2u\sin 30^\circ}{10} = \frac{u}{10}$
$u = 80\text{ m s}^{-1}$

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Q72MHT CET 2023 · 13 MAY · SHIFT 1Medium

Time of Flight, Maximum Height & Range · Area covered by bullets

A large number of bullets are fired in all directions with same speed ' $U$ '. The maximum area on the ground on which the bullets will spread is

Show answer and solution

Answer: (b)

Farthest a bullet can land is the maximum range at $45^\circ$ : $R_{max} = \frac{u^2}{g}$ .
Bullets fired in all directions cover a circle of radius $R_{max}$ .
$A = \pi R_{max}^2 = \frac{\pi u^4}{g^2}$
So the answer is (b) $\frac{\pi u^4}{g^2}$ .

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Q73MHT CET 2022 · 11 AUG · SHIFT 2Easy

Time of Flight, Maximum Height & Range · Time of flight

A shell is fired at an angle of $30^\circ$ to the horizontal with velocity $196\text{ m/s}$ . The time of flight is $\left[\sin 30^\circ = \frac{1}{2} = \cos 60^\circ\right]$

Show answer and solution

Answer: (b)

$T = \frac{2u\sin\theta}{g}$
$T = \frac{2 \times 196 \times 0.5}{9.8} = 20\text{ s}$
So the answer is (b) 20 s.

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Q74NEET 2022 · PHASE 2Easy

Time of Flight, Maximum Height & Range · Maximum height

A cricket ball is thrown by a player at a speed of 20 m/s in a direction 30 $^\circ$ above the horizontal. The maximum height attained by the ball during its motion is

(g = 10 m/s2)

Show answer and solution

Answer: (b)

$H = \frac{u^2\sin^2\theta}{2g}$
$H = \frac{400 \times \frac{1}{4}}{20} = 5\text{ m}$
So the answer is (b) 5 m.

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Q75JEE Main 2022 · 29 JUN · SHIFT 2Easy

Time of Flight, Maximum Height & Range · Range & maximum range

A person can throw a ball upto a maximum range of 100 m. How high above the ground he can throw the same ball?

Show answer and solution

Answer: (b)

$R_{max} = \frac{u^2}{g} = 100\text{ m}$
Maximum height (thrown vertically): $\frac{u^2}{2g} = \frac{100}{2} = 50\text{ m}$
So the answer is (b) 50 m.

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Q76JEE Main 2022 · 25 JUL · SHIFT 2Easy

Time of Flight, Maximum Height & Range · Relation between R and H

A ball is projected from the ground with a speed 15 ms ${}^{-1}$ at an angle $\theta$ with horizontal so that its range and maximum height are equal,

then 'tan $\theta$ ' will be equal to:

Show answer and solution

Answer: (d)

$\frac{R}{H} = \frac{4}{\tan\theta}$ (independent of speed)
$R = H \Rightarrow \tan\theta = 4$
So the answer is (d) 4.

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Q77JEE Main 2022 · 24 JUN · SHIFT 1Medium

Time of Flight, Maximum Height & Range · Time of flight

A projectile is projected with velocity of 25 m/s at an angle $\theta$ with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of $\theta$ will be:

[use g = 10 m/s2]

Show answer and solution

Answer: (d)

Inclination zero means the top: $t = \frac{u\sin\theta}{g} \Rightarrow u\sin\theta = 10t$
$R = \frac{2(u\sin\theta)(u\cos\theta)}{g} = 2t\,u\cos\theta \Rightarrow u\cos\theta = \frac{R}{2t}$
$\tan\theta = \frac{10t}{R/2t} = \frac{20t^2}{R} \Rightarrow \cot\theta = \frac{R}{20t^2}$
So the answer is (d) $\cot^{-1}\left(\frac{R}{20t^2}\right)$ .

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Q78MHT CET 2021 · 20 SEP · SHIFT 2Easy

Time of Flight, Maximum Height & Range · Time of flight

A cricket player hit a ball like a projectile but the fielder caught the ball after 2 second. The maximum height reached by a ball is (g = 10 m/s2)

Show answer and solution

Answer: (b)

$T = \frac{2u\sin\theta}{g}$ and $H = \frac{u^2\sin^2\theta}{2g}$ give $H = \frac{gT^2}{8}$ .
$H = \frac{10 \times 2^2}{8} = 5\text{ m}$
So the answer is (b) 5 m.

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Q79JEE Main 2021 · 27 AUG · SHIFT 2Easy

Time of Flight, Maximum Height & Range · Time of flight

A player kicks a football with an initial speed of 25 ms ${}^{-1}$ at an angle of 45 $^\circ$ from the ground. What are the maximum height and the time taken by the football to reach at the highest point during motion? (Take g = 10 ms ${}^{-2}$ )

Show answer and solution

Answer: (c)

$H = \frac{u^2\sin^2\theta}{2g} = \frac{625 \times \frac{1}{2}}{20} = 15.625\text{ m}$
Time to reach the top: $t = \frac{u\sin\theta}{g} = \frac{25 \times 0.707}{10} \approx 1.77\text{ s}$
So $h_{max} = 15.625\text{ m}$ , $t = 1.77\text{ s}$ , answer (c).

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Q80NEET 2021Medium

Time of Flight, Maximum Height & Range · Maximum height

A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution.

If this particle were projected with the same speed at an angle ' $\theta$ ' to the horizontal, the maximum height attained by it equals 4R. The angle of projection, $\theta$ , is then given by:

Show answer and solution

Answer: (a)

Speed in the circle: $v = \frac{2\pi R}{T}$
$H = \frac{v^2\sin^2\theta}{2g} = 4R \Rightarrow \sin^2\theta = \frac{8gR}{v^2} = \frac{8gR\,T^2}{4\pi^2R^2}$
$\sin^2\theta = \frac{2gT^2}{\pi^2 R}$
So the answer is (a) $\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2 R}\right)^{1/2}$ .

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Q81JEE Main 2019 · 10 JAN · SHIFT 1Medium

Time of Flight, Maximum Height & Range · Area covered by bullets

Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is -

Show answer and solution

Answer: (a)

Maximum area $A = \pi R_{max}^2 = \frac{\pi u^4}{g^2}$ , so $A \propto u^4$ .
$\frac{A_A}{A_B} = \left(\frac{1}{2}\right)^4 = \frac{1}{16}$
So the answer is (a) $1 : 16$ .

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Q82JEE Main 2014 · 11 APR · SHIFT 1Medium

Time of Flight, Maximum Height & Range · Time of flight

The initial speed of a bullet fired from a rifle is 630 m/s. The rifle is fired at the center of a target 700 m away at the same level as the target. How far above the center of the target the rifle must be aimed in order to hit the target?

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Answer: (c)

Time to cover 700 m horizontally: $t = \frac{700}{630} = \frac{10}{9}\text{ s}$ (angle is tiny, so $u_x \approx u$ )
Vertical drop in this time: $h = \frac{1}{2}gt^2 = 4.9 \times \frac{100}{81} \approx 6.05\text{ m}$
Aim this much above the centre: about 6.1 m, answer (c).

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Q83NEET 2014 · AIPMTMedium

Time of Flight, Maximum Height & Range · Maximum height

A projectile is fired from the surface of the earth with a velocity of 5 m s−1 and angle $\theta$ with the horizontal. Another projectile fired from another planet with a velocity of 3 m s−1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in m s ${}^{-2}$ ) is

(Given g = 9.8 m s ${}^{-2}$ )

Show answer and solution

Answer: (a)

Identical trajectory at the same angle needs the same $\frac{u^2}{g}$ (range and height both $\propto \frac{u^2}{g}$ ).
$\frac{5^2}{9.8} = \frac{3^2}{g'} \Rightarrow g' = 9.8 \times \frac{9}{25}$
$g' \approx 3.5\text{ m s}^{-2}$
So the answer is (a) 3.5.

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Q84JEE Main 2012 · AIEEEEasy

Time of Flight, Maximum Height & Range · Range & maximum range

A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be

Show answer and solution

Answer: (d)

Maximum height (vertical throw): $\frac{u^2}{2g} = 10\text{ m}$
$R_{max} = \frac{u^2}{g} = 2 \times 10 = 20\text{ m}$
So the answer is (d) 20 m.

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Q85NEET 2012 · AIPMT · PRELIMSEasy

Time of Flight, Maximum Height & Range · Relation between R and H

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is

Show answer and solution

Answer: (b)

$\frac{R}{H} = \frac{4}{\tan\theta}$
$R = H \Rightarrow \tan\theta = 4$
So the answer is (b) $\theta = \tan^{-1}(4)$ .

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Q86NEET 2011 · AIPMT · PRELIMSEasy

Time of Flight, Maximum Height & Range · Range & maximum range

A missile is fired for maximum range with an initial velocity of 20 m/s. If g = 10 m/s2, the range of the missile is

Show answer and solution

Answer: (a)

Maximum range at $45^\circ$ : $R_{max} = \frac{u^2}{g}$
$R_{max} = \frac{400}{10} = 40\text{ m}$
So the answer is (a) 40 m.

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Q87JEE Main 2011 · AIEEEMedium

Time of Flight, Maximum Height & Range · Area covered by bullets

A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, the total area around the fountain that gets wet is:

Show answer and solution

Answer: (a)

Water travels farthest at $45^\circ$ : $R_{max} = \frac{v^2}{g}$ .
Wet region is a circle of radius $R_{max}$ : $A = \pi\left(\frac{v^2}{g}\right)^2$
So the answer is (a) $\pi\frac{v^4}{g^2}$ .

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Q88NEET 1990 · AIPMTEasy

Time of Flight, Maximum Height & Range · Range & maximum range

The maximum range of a gun of horizontal terrain is 16 km. If $g = 10\text{ m s}^{-2}$ , then muzzle velocity of a shell must be

Show answer and solution

Answer: (c)

$R_{max} = \frac{u^2}{g} \Rightarrow u^2 = 16000 \times 10 = 1.6 \times 10^5$
$u = 400\text{ m s}^{-1}$
So the answer is (c).

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Q89MHT CET 2026 · 16 APR · SHIFT 2Easy

Comparing Two Projectiles · Ratios of R, H and T

Two objects are projected with same velocity 'u' at different angles $\alpha$ and $\beta$ with the horizontal. If $\alpha + \beta = 90^\circ$ , the ratio of horizontal range of the first object to the second object will be $[\sin(\pi - \theta) = \sin\theta]$

Show answer and solution

Answer: (c)

$R = \frac{u^2\sin 2\theta}{g}$ ; with $\beta = 90^\circ - \alpha$ , $\sin 2\beta = \sin(180^\circ - 2\alpha) = \sin 2\alpha$ .
So $R_1 = R_2$ .
So the answer is (c) $1 : 1$ .

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Q90JEE Main 2026 · 4 APR · SHIFT 1Easy

Comparing Two Projectiles · Ratios of R, H and T

The two projectiles are projected with the same initial velocities at the $15^\circ$ and $30^\circ$ with respect to the horizontal. The ratio of their ranges is $1 : x$ . The value of $x$ is

Show answer and solution

Answer: (b)

$R \propto \sin 2\theta$ for the same speed.
$\frac{R_1}{R_2} = \frac{\sin 30^\circ}{\sin 60^\circ} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$
So $x = \sqrt{3}$ , answer (b).

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Q91JEE Main 2026 · 8 APR · SHIFT 2Medium

Comparing Two Projectiles · Complementary angles

Two identical bodies, projected with the same speed at two different angles cover the same horizontal range $R$ . If the time of flight of these bodies are 5 s and 10 s, respectively, then the value of $R$ is

______ m. (Take $g = 10\text{ m/s}^{2}$ )

Show answer and solution

Answer: (a)

Same range at the same speed: angles $\theta$ and $90^\circ - \theta$ .
$T_1 T_2 = \frac{2u\sin\theta}{g}\cdot\frac{2u\cos\theta}{g} = \frac{2}{g}\cdot\frac{u^2\sin 2\theta}{g} = \frac{2R}{g}$
$5 \times 10 = \frac{2R}{10} \Rightarrow R = 250\text{ m}$
So the answer is (a) 250.

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Q92MHT CET 2026 · 13 MAY · SHIFT 2Medium

Comparing Two Projectiles · Same height or same time

If two spheres are projected at angles $45^\circ$ and $60^\circ$ with horizontal and the maximum heights reached are same. The ratio of their initial velocities is ( $\sin 45^\circ = \frac{1}{\sqrt{2}}, \sin 60^\circ = \frac{\sqrt{3}}{2}$ )

Show answer and solution

Answer: (d)

Equal $H = \frac{u^2\sin^2\theta}{2g}$ means $u_1\sin 45^\circ = u_2\sin 60^\circ$ .
$\frac{u_1}{u_2} = \frac{\sin 60^\circ}{\sin 45^\circ} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}$
So the answer is (d) $\sqrt{3} : \sqrt{2}$ .

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Q93MHT CET 2026 · 20 APR · SHIFT 2Medium

Comparing Two Projectiles · Same height or same time

Two spheres are projected at angles $30^\circ$ and $45^\circ$ with the horizontal. The maximum height reached by both is same. The ratio of their initial velocities is, $\left(\sin 45^\circ = \frac{1}{\sqrt{2}}, \sin 30^\circ = 0.5\right)$

Show answer and solution

Answer: (b)

Equal $H = \frac{u^2\sin^2\theta}{2g}$ means $u_1\sin 30^\circ = u_2\sin 45^\circ$ .
$\frac{u_1}{u_2} = \frac{\sin 45^\circ}{\sin 30^\circ} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$
So the answer is (b) $\sqrt{2} : 1$ .

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Q94MHT CET 2026 · 17 APR · SHIFT 1Medium

Comparing Two Projectiles · Same height or same time

Two balls A and B are projected at an angle of $45^\circ$ and $60^\circ$ respectively, so that the maximum heights reached are same for both. The ratio of initial velocity of projection of ball A to that for ball B is

$\left(\sin 30^\circ = \cos 60^\circ = \frac{1}{2},\ \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}},\ \sin 60^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2}\right)$

Show answer and solution

Answer: (d)

Equal $H = \frac{u^2\sin^2\theta}{2g}$ means $u_A\sin 45^\circ = u_B\sin 60^\circ$ .
$\frac{u_A}{u_B} = \frac{\sin 60^\circ}{\sin 45^\circ} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \frac{\sqrt{3}}{\sqrt{2}}$
So the answer is (d) $\sqrt{3} : \sqrt{2}$ .

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Q95JEE Main 2025 · 29 JAN · SHIFT 1Easy

Comparing Two Projectiles · Same height or same time

Two projectiles are fired with same initial speed from same point on ground at angles of $(45^\circ - \alpha)$ and $(45^\circ + \alpha)$ , respectively, with the horizontal direction. The ratio of their maximum heights attained is:

Show answer and solution

Answer: (d)

$H \propto \sin^2\theta$ , so $\frac{H_1}{H_2} = \frac{\sin^2(45^\circ - \alpha)}{\sin^2(45^\circ + \alpha)}$
$\sin^2(45^\circ \mp \alpha) = \frac{1 - \cos(90^\circ \mp 2\alpha)}{2} = \frac{1 \mp \sin 2\alpha}{2}$
$\frac{H_1}{H_2} = \frac{1 - \sin 2\alpha}{1 + \sin 2\alpha}$
So the answer is (d).

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Q96JEE Main 2025 · 7 APR · SHIFT 1Easy

Comparing Two Projectiles · Ratios of R, H and T

Two projectiles are fired from ground with same initial speeds from same point at angles $(45^\circ + \alpha)$ and $(45^\circ - \alpha)$ with horizontal direction. The ratio of their times of flights is

Show answer and solution

Answer: (a)

$T = \frac{2u\sin\theta}{g}$ , so $\frac{T_1}{T_2} = \frac{\sin(45^\circ + \alpha)}{\sin(45^\circ - \alpha)}$
$= \frac{\cos\alpha + \sin\alpha}{\cos\alpha - \sin\alpha}$ (the $\frac{1}{\sqrt{2}}$ cancels)
Divide by $\cos\alpha$ : $\frac{1 + \tan\alpha}{1 - \tan\alpha}$
So the answer is (a).

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Q97JEE Main 2025 · 8 APR · SHIFT 2Medium

Comparing Two Projectiles · Ratios of R, H and T

Two balls with same mass and initial velocity, are projected at different angles in such a way that maximum height reached by first ball is 8 times higher than that of the second ball. $T_1$ and $T_2$ are the total flying times of first and second ball, respectively, then the ratio of $T_1$ and $T_2$ is

Show answer and solution

Answer: (c)

Same speed: $H \propto \sin^2\theta$ and $T \propto \sin\theta$ , so $T \propto \sqrt{H}$ .
$\frac{T_1}{T_2} = \sqrt{\frac{H_1}{H_2}} = \sqrt{8} = 2\sqrt{2}$
So the answer is (c) $2\sqrt{2} : 1$ .

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Q98JEE Main 2024 · 30 JAN · SHIFT 2Difficult

Comparing Two Projectiles · Ratios of R, H and T

Projectiles A and B are thrown at angles of $45^\circ$ and $60^\circ$ with vertical respectively from top of a $400\text{ m}$ high tower. If their ranges and times of flight are same, the ratio of their speeds of projection $v_A : v_B$ is:

[Take $g = 10\text{ m s}^{-2}$ ]

Show answer and solution

Answer: (b)

From the same height, equal times of flight need equal vertical components: $v_A\cos 45^\circ = v_B\cos 60^\circ$ (angles from vertical).
$\frac{v_A}{v_B} = \frac{\cos 60^\circ}{\cos 45^\circ} = \frac{1/2}{1/\sqrt{2}} = \frac{1}{\sqrt{2}}$
Equal ranges with equal times would also need $v_A\sin 45^\circ = v_B\sin 60^\circ$ , i.e. $\sqrt{3} : \sqrt{2}$ , so both conditions cannot hold together.
Using the time-of-flight condition, $v_A : v_B = 1 : \sqrt{2}$ , option (c).

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Q99JEE Main 2023 · 31 JAN · SHIFT 2Easy

Comparing Two Projectiles · Complementary angles

Two bodies are projected from ground with same speeds $40\text{ m s}^{-1}$ at two different angles with respect to horizontal. The bodies were found to have same range. If one of the body was projected at an angle of $60^\circ$ , with horizontal then sum of the maximum heights, attained by the two projectiles, is ________ $\text{m}$ . (Given $g = 10\text{ m s}^{-2}$ )

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 80

Same range at the same speed means complementary angles: $60^\circ$ and $30^\circ$ .
$H_1 + H_2 = \frac{u^2(\sin^2 60^\circ + \sin^2 30^\circ)}{2g} = \frac{u^2}{2g}$
$= \frac{40^2}{2\times 10} = 80\text{ m}$

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Q100JEE Main 2023 · 10 APR · SHIFT 2Easy

Comparing Two Projectiles · Same height or same time

Two projectiles are projected at $30^\circ$ and $60^\circ$ with the horizontal with the same speed. The ratio of the maximum height attained by the two projectiles respectively is:

Show answer and solution

Answer: (c)

$H = \frac{u^2\sin^2\theta}{2g}$ , so $H \propto \sin^2\theta$ for the same speed.
$\frac{H_1}{H_2} = \frac{\sin^2 30^\circ}{\sin^2 60^\circ} = \frac{1/4}{3/4} = \frac{1}{3}$
So the answer is (c) $1 : 3$ .

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Q101JEE Main 2023 · 10 APR · SHIFT 1Easy

Comparing Two Projectiles · Ratios of R, H and T

The range of the projectile projected at an angle of 15 $^\circ$ with horizontal is 50 m. If the projectile is projected with same velocity at an angle of 45 $^\circ$ with horizontal, then its range will be

Show answer and solution

Answer: (b)

$R = \frac{u^2\sin 2\theta}{g}$ : at $15^\circ$ , $50 = \frac{u^2}{g}\sin 30^\circ \Rightarrow \frac{u^2}{g} = 100\text{ m}$
At $45^\circ$ : $R = \frac{u^2}{g}\sin 90^\circ = 100\text{ m}$
So the answer is (b) 100 m.

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Q102JEE Main 2023 · 25 JAN · SHIFT 2Easy

Comparing Two Projectiles · Ratios of R, H and T

Two objects are projected with same velocity 'u' however at different angles $\alpha$ and $\beta$ with the horizontal. If $\alpha + \beta = 90^\circ$ , the ratio of horizontal range of the first object to the 2nd object will be:

Show answer and solution

Answer: (a)

$\beta = 90^\circ - \alpha \Rightarrow \sin 2\beta = \sin(180^\circ - 2\alpha) = \sin 2\alpha$
Same $u$ , so $R_1 = R_2$ .
So the answer is (a) $1 : 1$ .

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Q103JEE Main 2023 · 8 APR · SHIFT 1Medium

Comparing Two Projectiles · Ratios of R, H and T

Two projectiles A and B are thrown with initial velocities of $40\text{ m/s}$ and $60\text{ m/s}$ at angles $30^\circ$ and $60^\circ$ with the horizontal respectively. The ratio of their ranges respectively is $\left(g = 10\text{ m/s}^{2}\right)$

Show answer and solution

Answer: (a)

$R = \frac{u^2\sin 2\theta}{g}$
$\frac{R_A}{R_B} = \frac{40^2\sin 60^\circ}{60^2\sin 120^\circ} = \frac{1600}{3600}$ (as $\sin 60^\circ = \sin 120^\circ$ )
$= \frac{4}{9}$ , so the answer is (a) $4 : 9$ .

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Q104JEE Main 2022 · 26 JUL · SHIFT 2Easy

Comparing Two Projectiles · Ratios of R, H and T

Two projectiles are thrown with same initial velocity making an angle of $45^\circ$ and $30^\circ$ with the horizontal respectively. The ratio of their respective ranges will be:

Show answer and solution

Answer: (c)

$R \propto \sin 2\theta$ for the same speed.
$\frac{R_1}{R_2} = \frac{\sin 90^\circ}{\sin 60^\circ} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}}$
So the answer is (c) $2 : \sqrt{3}$ .

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Q105JEE Main 2022 · 25 JUN · SHIFT 2Medium

Comparing Two Projectiles · Complementary angles

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Two identical balls A and B thrown with same velocity 'u' at two different angles with horizontal attained the same range R. If A and B reached the maximum height h1 and h2 respectively, then $R = 4\sqrt{h_1 h_2}$

Reason R: Product of said heights.

$h_1 h_2 = \left(\frac{u^2 \sin^2\theta}{2g}\right) \cdot \left(\frac{u^2 \cos^2\theta}{2g}\right)$

Choose the correct answer:

Show answer and solution

Answer: (a)

Same range means angles $\theta$ and $90^\circ - \theta$ , so $h_1 = \frac{u^2\sin^2\theta}{2g}$ , $h_2 = \frac{u^2\cos^2\theta}{2g}$ : R is true.
$h_1 h_2 = \frac{u^4\sin^2\theta\cos^2\theta}{4g^2}$ and $R^2 = \frac{4u^4\sin^2\theta\cos^2\theta}{g^2} = 16h_1 h_2$
So $R = 4\sqrt{h_1 h_2}$ : A is true and follows directly from R.
So the answer is (a).

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Q106JEE Main 2022 · 26 JUL · SHIFT 1Medium

Comparing Two Projectiles · Same height or same time

Two projectiles thrown at $30^\circ$ and $45^\circ$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is:

Show answer and solution

Answer: (c)

Time to maximum height $t = \frac{u\sin\theta}{g}$ ; equal times give $u_1\sin 30^\circ = u_2\sin 45^\circ$ .
$\frac{u_1}{u_2} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$
So the answer is (c) $\sqrt{2} : 1$ .

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Q107JEE Main 2022 · 30 JUN · SHIFT 1Medium

Comparing Two Projectiles · Same height or same time

Two projectiles P1 and P2 thrown with speed in the ratio $\sqrt{3}$ : $\sqrt{2}$ , attain the same height during their motion. If P2 is thrown at an angle of 60 $^\circ$ with the horizontal, the angle of projection of P1 with horizontal will be:

Show answer and solution

Answer: (c)

Same maximum height means equal vertical components: $u_1\sin\theta_1 = u_2\sin 60^\circ$ .
$\sin\theta_1 = \frac{\sqrt{2}}{\sqrt{3}}\cdot\frac{\sqrt{3}}{2} = \frac{1}{\sqrt{2}}$
So $\theta_1 = 45^\circ$ , answer (c).

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Q108JEE Main 2022 · 27 JUL · SHIFT 1Medium

Comparing Two Projectiles · Same height or same time

A ball of mass m is thrown vertically upward. Another ball of mass $2m$ is thrown at an angle $\theta$ with the vertical. Both the balls stay in air for the same period of time. The ratio of the heights attained by the two balls respectively is $\frac{1}{x}$ . The value of x is _____________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 1

Time of flight $T = \frac{2u_y}{g}$ depends only on the vertical component $u_y$ (not on mass).
Same $T$ means same $u_y$ , so same height $H = \frac{u_y^2}{2g}$ .
Ratio $\frac{1}{x} = 1 \Rightarrow x = 1$

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Q109JEE Main 2022 · 30 JUL · SHIFT 2Medium

Comparing Two Projectiles · Ratios of R, H and T

Two projectiles are thrown towards each other at $15^\circ$ and $45^\circ$ angles respectively with the horizontal at the same speed. The difference between the horizontal distances traveled by the two projectiles is $80\text{ m}$ . The initial speed of the projectiles is ______ $\text{m s}^{-1}$

[Given $g = 10\text{ m s}^{-2}$ ]

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 40

$R_{45} - R_{15} = \frac{u^2}{g}(\sin 90^\circ - \sin 30^\circ) = \frac{u^2}{2g}$
$\frac{u^2}{2\times 10} = 80 \Rightarrow u^2 = 1600$
$u = 40\text{ m s}^{-1}$

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Q110JEE Main 2022 · 29 JUL · SHIFT 1Medium

Comparing Two Projectiles · Ratios of R, H and T

An object is projected in the air with initial velocity u at an angle $\theta$ . The projectile motion is such that the horizontal range R, is maximum. Another object is projected in the air with a horizontal range half of the range of first object. The initial velocity remains same in both the cases. The value of the angle of projection, at which the second object is projected, will be _________ degree.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 15

Maximum range at $45^\circ$ : $R = \frac{u^2}{g}$ .
Second object: $\frac{u^2\sin 2\theta}{g} = \frac{u^2}{2g} \Rightarrow \sin 2\theta = \frac{1}{2}$
$2\theta = 30^\circ \Rightarrow \theta = 15^\circ$

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Q111JEE Main 2021 · 1 SEP · SHIFT 2Easy

Comparing Two Projectiles · Complementary angles

The ranges and heights for two projectiles projected with the same initial velocity at angles 42 $^\circ$ and 48 $^\circ$ with the horizontal are R1, R2 and H1, H2 respectively. Choose the correct option:

Show answer and solution

Answer: (b)

$42^\circ$ and $48^\circ$ are complementary, so $\sin 84^\circ = \sin 96^\circ$ gives $R_1 = R_2$ .
$H = \frac{u^2\sin^2\theta}{2g}$ grows with $\theta$ , so $H_1 < H_2$ .
So the answer is (b) $R_1 = R_2$ and $H_1 < H_2$ .

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Q112JEE Main 2019 · 12 APR · SHIFT 2Medium

Comparing Two Projectiles · Complementary angles

Two particles are projected from the same point with the same speed u such that they have the same range R, but different maximum heights, h1 and h2. Which of the following is correct?

Show answer and solution

Answer: (b)

Same range: angles $\theta$ and $90^\circ - \theta$ , so $h_1 = \frac{u^2\sin^2\theta}{2g}$ , $h_2 = \frac{u^2\cos^2\theta}{2g}$
$h_1 h_2 = \frac{u^4\sin^2\theta\cos^2\theta}{4g^2}$ , while $R^2 = \frac{4u^4\sin^2\theta\cos^2\theta}{g^2}$
$R^2 = 16h_1 h_2$
So the answer is (b).

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Q113JEE Main 2019 · 12 APR · SHIFT 1Medium

Comparing Two Projectiles · Complementary angles

A shell is fired from a fixed artillery gun with an initial speed u such that it hits the target on the ground at a distance R from it. If t1 and t2 are the values of the time taken by it to hit the target in two possible ways, the product t1t2 is -

Show answer and solution

Answer: (a)

The two ways are angles $\theta$ and $90^\circ - \theta$ : $t_1 = \frac{2u\sin\theta}{g}$ , $t_2 = \frac{2u\cos\theta}{g}$
$t_1 t_2 = \frac{2}{g}\cdot\frac{2u^2\sin\theta\cos\theta}{g} = \frac{2R}{g}$
So the answer is (a) $\frac{2R}{g}$ .

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Q114NEET 2006 · AIPMTEasy

Comparing Two Projectiles · Ratios of R, H and T

For angles of projection of a projectile at angle (45o $-\theta$ ) and (45o + $\theta$ ), the horizontal range described by the projectile are in the ratio of

Show answer and solution

Answer: (b)

$45^\circ - \theta$ and $45^\circ + \theta$ add up to $90^\circ$ (complementary).
$\sin(90^\circ - 2\theta) = \sin(90^\circ + 2\theta) = \cos 2\theta$ , so the ranges are equal.
So the answer is (b) $1 : 1$ .

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Q115JEE Main 2005 · AIEEEMedium

Comparing Two Projectiles · Complementary angles

A projectile can have the same range $R$ for two angles of projection. If $t_1$ and $t_2$ be the times of flights in the two cases, then the product of the two time of flights is proportional to

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Answer: (d)

For angles $\theta$ and $90^\circ - \theta$ : $t_1 t_2 = \frac{2u\sin\theta}{g}\cdot\frac{2u\cos\theta}{g}$
$= \frac{2}{g}\cdot\frac{u^2\sin 2\theta}{g} = \frac{2R}{g}$
So $t_1 t_2 \propto R$ , answer (d).

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Q116JEE Main 2004 · AIEEEMedium

Comparing Two Projectiles · Complementary angles

A projectile can have the same range 'R' for two angles of projection. If T1 and T2 be the time of flights in the two cases, then the product of the two time of flights is directly proportional to

Show answer and solution

Answer: (a)

For angles $\theta$ and $90^\circ - \theta$ : $T_1 T_2 = \frac{2u\sin\theta}{g}\cdot\frac{2u\cos\theta}{g}$
$= \frac{2}{g}\cdot\frac{u^2\sin 2\theta}{g} = \frac{2R}{g}$
So $T_1 T_2 \propto R$ , answer (a).

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Q117NEET 2000 · AIPMTEasy

Comparing Two Projectiles · Ratios of R, H and T

Two projectiles of same mass and with same velocity are thrown at an angle 60o and 30o with the horizontal, then which will remain same

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Answer: (b)

$60^\circ$ and $30^\circ$ are complementary: $\sin 120^\circ = \sin 60^\circ$ , so the range is the same.
$T \propto \sin\theta$ and $H \propto \sin^2\theta$ differ for the two angles.
So the answer is (b) range of projectile.

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Q118NEET 1992 · AIPMTEasy

Comparing Two Projectiles · Ratios of R, H and T

If a body $A$ of mass $M$ is thrown with velocity $v$ at an angle of $30^\circ$ to the horizontal and another body $B$ of the same mass is thrown with the same speed at an angle of $60^\circ$ to the horizontal, the ratio of horizontal range of $A$ to $B$ will be

Show answer and solution

Answer: (b)

$R = \frac{v^2\sin 2\theta}{g}$ , independent of mass.
$\frac{R_A}{R_B} = \frac{\sin 60^\circ}{\sin 120^\circ} = 1$
So the answer is (b) $1 : 1$ .

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Q119NEET 1990 · AIPMTEasy

Comparing Two Projectiles · Ratios of R, H and T

If a body $A$ of mass $M$ is thrown with velocity $v$ at an angle of $30^\circ$ to the horizontal and another body $B$ of the same mass is thrown with the same speed at an angle of $60^\circ$ to the horizontal, the ratio of horizontal range of $A$ to $B$ will be

Show answer and solution

Answer: (b)

$R = \frac{v^2\sin 2\theta}{g}$ , independent of mass.
$\frac{R_A}{R_B} = \frac{\sin 60^\circ}{\sin 120^\circ} = 1$
So the answer is (b) $1 : 1$ .

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Q120JEE Main 2025 · 7 APR · SHIFT 2Medium

Horizontal Projectile & Inclined Plane · Bomb from an aeroplane

A helicopter flying horizontally with a speed of 360 km/h at an altitude of 2 km, drops an object at an instant. The object hits the ground at a point O, 20 s after it is dropped. Displacement of 'O' from the position of helicopter where the object was released is:

(use acceleration due to gravity g = 10 m/s2 and neglect air resistance)

Show answer and solution

Answer: (c)

$v = 360\text{ km/h} = 100\text{ m/s}$ ; check $t = \sqrt{\frac{2(2000)}{10}} = 20\text{ s}$ .
Horizontal distance $x = 100 \times 20 = 2000\text{ m} = 2\text{ km}$ ; vertical drop $= 2\text{ km}$ .
Displacement $= \sqrt{2^2 + 2^2} = 2\sqrt{2}\text{ km}$
So the answer is (c).

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Q121JEE Main 2024 · 8 APR · SHIFT 2Medium

Horizontal Projectile & Inclined Plane · Body thrown from a tower

A body of mass M thrown horizontally with velocity v from the top of the tower of height H touches the ground at a distance of $100\text{ m}$ from the foot of the tower. A body of mass $2M$ thrown at a velocity $\frac{v}{2}$ from the top of the tower of height $4H$ will touch the ground at a distance of _______ m.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 100

Horizontal range from a tower: $R = v\sqrt{\frac{2H}{g}}$ (independent of mass).
New: $R' = \frac{v}{2}\sqrt{\frac{2(4H)}{g}} = \frac{v}{2} \cdot 2\sqrt{\frac{2H}{g}} = R$
$R' = 100\text{ m}$

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Q122JEE Main 2024 · 29 JAN · SHIFT 1Difficult

Horizontal Projectile & Inclined Plane · Body thrown from a tower

A ball rolls off the top of a stairway with horizontal velocity $u$ . The steps are $0.1\text{ m}$ high and $0.1\text{ m}$ wide. The minimum velocity $u$ with which that ball just hits the step 5 of the stairway will be $\sqrt{x}\text{ m s}^{-1}$ where $x =$ __________ [use $g = 10\text{ m/s}^{2}$ ].

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2

To land on step 5 the ball must just clear the edge of step 4: fall $y = 4(0.1) = 0.4\text{ m}$ with $x = 0.4\text{ m}$ .
$t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{0.8}{10}} = \sqrt{0.08}\text{ s}$
$u = \frac{x}{t} = \frac{0.4}{\sqrt{0.08}} = \sqrt{\frac{0.16}{0.08}} = \sqrt{2}\text{ m/s}$
$x = 2$

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Q123JEE Main 2023 · 1 FEB · SHIFT 1Easy

Horizontal Projectile & Inclined Plane · Body thrown from a tower

A child stands on the edge of the cliff $10\text{ m}$ above the ground and throws a stone horizontally with an initial speed of $5\text{ m s}^{-1}$ . Neglecting the air resistance, the speed with which the stone hits the ground will be ________ $\text{m s}^{-1}$ (given, $g = 10\text{ m s}^{-2}$ ).

Show answer and solution

Answer: (d)

Horizontal: $v_x = 5\text{ m/s}$ stays constant.
Vertical: $v_y^2 = 2gh = 2(10)(10) = 200$
$v = \sqrt{v_x^2 + v_y^2} = \sqrt{25 + 200} = 15\text{ m/s}$
So the answer is (d) 15.

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Q124JEE Main 2022 · 30 JUN · SHIFT 1Medium

Horizontal Projectile & Inclined Plane · Release from a moving vehicle

At t = 0, truck, starting from rest, moves in the positive x-direction at uniform acceleration of 5 ms ${}^{-2}$ . At t = 20 s, a ball is released from the top of the truck. The ball strikes the ground in 1 s after the release. The velocity of the ball, when it strikes the ground, will be:

(Given g = 10 ms ${}^{-2}$ )

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Answer: (a)

At $t = 20\text{ s}$ : truck speed $= 5 \times 20 = 100\text{ m/s}$ , ball keeps $v_x = 100$ .
After release only gravity acts: in $1\text{ s}$ , $v_y = -10 \times 1 = -10\text{ m/s}$ .
$\vec{v} = 100\hat{i} - 10\hat{j}$
So the answer is (a).

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Q125MHT CET 2021 · 21 SEP · SHIFT 1Easy

Horizontal Projectile & Inclined Plane · Bomb from an aeroplane

A bomb is dropped by an aeroplane flying horizontally with a velocity $200\text{ km/hr}$ and at a height of $980\text{ m}$ . At the time of dropping a bomb, the distance of the aeroplane from the target on the ground to hit directly is $\left(g = 9.8\text{ m/s}^{2}\right)$

Show answer and solution

Answer: (c)

$v = 200\text{ km/h} = 200\times\frac{5}{18} = \frac{500}{9}\text{ m/s}$
Fall time $t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times 980}{9.8}} = \sqrt{200} = 10\sqrt{2}\text{ s}$
Horizontal distance $x = vt = \frac{500}{9}\times 10\sqrt{2} = \frac{5000\sqrt{2}}{9} = \frac{10^4}{9\sqrt{2}}\text{ m}$
So the answer is (c).

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Q126JEE Main 2021 · 31 AUG · SHIFT 1Easy

Horizontal Projectile & Inclined Plane · Bomb from an aeroplane

A helicopter is flying horizontally with a speed 'v' at an altitude 'h' has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped?

Show answer and solution

Answer: (c)

Time of fall $t = \sqrt{\frac{2h}{g}}$ ; in this time the packet moves $x = vt = v\sqrt{\frac{2h}{g}}$ horizontally.
The man is at horizontal distance $x$ and vertical distance $h$ from the helicopter.
Distance $= \sqrt{x^2 + h^2} = \sqrt{\frac{2v^2h}{g} + h^2}$
So the answer is (c).

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Q127NEET 2021Medium

Horizontal Projectile & Inclined Plane · Release from a moving vehicle

A car starts from rest and accelerates at 5m/s2. At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at t = 6 s? (Take g = 10 m/s2)

Show answer and solution

Answer: (a)

At $t = 4\text{ s}$ : car speed $= 5 \times 4 = 20\text{ m/s}$ , which the ball keeps horizontally.
After release only gravity acts: in $2\text{ s}$ , $v_y = 10 \times 2 = 20\text{ m/s}$ .
$v = \sqrt{20^2 + 20^2} = 20\sqrt{2}\text{ m/s}$ ; acceleration $= g = 10\text{ m/s}^2$
So the answer is (a).

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Q128NEET 2019Medium

Horizontal Projectile & Inclined Plane · Body thrown from a tower

Two bullets are fired horizontally and simultaneously towards each other from roof tops of two buildings 100 m apart and of same height of 200 m, with the same velocity of 25 m/s. When and where will the two bullets collide? ( $g = 10\text{ m/s}^{2}$ )

Show answer and solution

Answer: (b)

Relative horizontal speed $= 25 + 25 = 50\text{ m/s}$ , so they meet at $t = \frac{100}{50} = 2\text{ s}$ .
Both fall the same distance $\frac{1}{2}gt^2 = \frac{1}{2}(10)(2)^2 = 20\text{ m}$ , so they do collide.
Height $= 200 - 20 = 180\text{ m}$
So the answer is (b).

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Q129JEE Main 2019 · 10 APR · SHIFT 2Difficult

Horizontal Projectile & Inclined Plane · Projectile on an inclined plane

A plane is inclined at an angle $\alpha$ = 30° with respect to the horizontal. A particle is projected with a speed u = 2 ms–1, from the base of the plane, making an angle $\theta$ = 15° with respect to the plane as shown in the figure. The distance from the base, at which the particle hits the plane is close to: (Take g = 10 ms–2)

Show answer and solution

Answer: (c)

Range up an incline: $R = \frac{2u^2 \sin\theta \cos(\theta + \alpha)}{g\cos^2\alpha}$
$R = \frac{2(4)\sin 15^\circ \cos 45^\circ}{10 \times \frac{3}{4}} = \frac{8(0.259)(0.707)}{7.5}$
$R \approx 0.195\text{ m} \approx 20\text{ cm}$
So the answer is (c) 20 cm.

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Q130NEET 2002 · AIPMTEasy

Horizontal Projectile & Inclined Plane · Body thrown from a tower

A particle A is dropped from a height and another particle B is projected in horizontal direction with speed of 5 m/sec from the same height then correct statement is

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Answer: (c)

Both start with zero vertical velocity from the same height.
Vertical motion is independent of horizontal velocity, so $t = \sqrt{\frac{2h}{g}}$ for both.
B lands with greater speed since it also has a horizontal $5\text{ m/s}$ .
So the answer is (c) both reach simultaneously.

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Q131NEET 2000 · AIPMTEasy

Horizontal Projectile & Inclined Plane · Body thrown from a tower

A solid sphere and a hollow sphere are thrown horizontally from a cliff with equal velocities, respectively. Then which sphere reaches first on earth?

Show answer and solution

Answer: (c)

Time of fall $t = \sqrt{\frac{2h}{g}}$ depends only on height and $g$ .
It does not depend on mass, shape or horizontal speed (no air resistance).
So the answer is (c) both reach simultaneously.

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Q132MHT CET 2026 · 20 APR · SHIFT 1Medium

Relative Velocity in Two Dimensions · River & boat

A boat crosses a river from one bank A to another bank B which is opposite. The distance between them is D. The speed of water is $V_W$ and that of boat relative to water is $V_B$ . If $V_B = 2V_W$ , the time taken by the boat to cross the river directly along AB is $\left(\sin 30^\circ = \frac{1}{2}\right) \cdot \left(\cos 30^\circ = \frac{\sqrt{3}}{2}\right)$

Show answer and solution

Answer: (c)

To go straight along AB, head upstream at $\phi$ from AB: $\sin\phi = \frac{V_W}{V_B} = \frac{1}{2} \Rightarrow \phi = 30^\circ$
Speed along AB $= V_B\cos 30^\circ = \frac{\sqrt{3}}{2}V_B$
$t = \frac{D}{\frac{\sqrt{3}}{2}V_B} = \frac{2D}{V_B\sqrt{3}}$
So the answer is (c).

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Q133JEE Main 2026 · 21 JAN · SHIFT 2Medium

Relative Velocity in Two Dimensions · River & boat

A river of width 200 m is flowing from west to east with a speed of 18 km/h. A boat, moving with speed of 36 km/h in still water, is made to travel one-round trip (bank to bank of the river). Minimum time taken by the boat for this journey and also the displacement along the river bank are ______ and ______ respectively.

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Answer: (c)

Minimum time: head perpendicular to flow. $v_b = 36\text{ km/h} = 10\text{ m/s}$ , $v_r = 18\text{ km/h} = 5\text{ m/s}$ .
One crossing: $t = \frac{200}{10} = 20\text{ s}$ ; round trip $= 40\text{ s}$ .
Drift each way $= 5 \times 20 = 100\text{ m}$ downstream; both crossings drift east, total $200\text{ m}$ .
So the answer is (c) 40 s and 200 m.

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Q134MHT CET 2026 · 16 APR · SHIFT 1Medium

Relative Velocity in Two Dimensions · Chasing & minimum distance

Two boys are standing at points A and B on ground where distance AB = a. The boy at point B starts running perpendicular to line AB with velocity ' $V_1$ '. The boy at point A starts running simultaneously with velocity 'V' and catches the other boy in time 't'. The value of 't' is

Show answer and solution

Answer: (a)

They meet at a point C with $BC = V_1 t$ , $AC = Vt$ and $AB \perp BC$ .
$(Vt)^2 = a^2 + (V_1 t)^2$
$t = \left[\frac{a^2}{V^2 - V_1^2}\right]^{1/2}$
So the answer is (a).

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Q135MHT CET 2026 · 12 MAY · SHIFT 1Difficult

Relative Velocity in Two Dimensions · Seen from a moving frame

Airplanes A and B are flying with constant velocity in the same vertical plane at angles $30^\circ$ and $60^\circ$ with respect to the horizontal as shown in the figure. The speed of A is $100\sqrt{3}$ m/s. At time $t = 0$ s, an observer in A finds B at a distance of $500$ m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at $t = t_0$ , A just escapes being hit by B, $t_0$ in second is

$\{\sin(30^\circ) = 0.5, \cos(30^\circ) = 3^{0.5}/2\}$

Show answer and solution

Answer: (d)

B is seen moving perpendicular to A's path, so the component of $\vec v_B$ along A's direction equals $v_A$ .
Angle between the two paths is $60^\circ - 30^\circ = 30^\circ$ : $v_B\cos 30^\circ = 100\sqrt{3} \Rightarrow v_B = 200$ m/s.
Relative speed (perpendicular to A's path) $= v_B\sin 30^\circ = 200 \times 0.5 = 100$ m/s.
$t_0 = \dfrac{500}{100} = 5$ s
So the answer is (d) $5$ .

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Q136JEE Main 2025 · 2 APR · SHIFT 1Medium

Relative Velocity in Two Dimensions · River & boat

A river is flowing from west to east direction with speed of $9\text{ km h}^{-1}$ . If a boat capable of moving at a maximum speed of $27\text{ km h}^{-1}$ in still water, crosses the river in half a minute, while moving with maximum speed at an angle of $150^\circ$ to direction of river flow, then the width of the river is:

Show answer and solution

Answer: (a)

Component of boat velocity across the river: $27\sin 150^\circ = 13.5\text{ km/h} = 3.75\text{ m/s}$
Width $= 3.75 \times 30 = 112.5\text{ m}$
So the answer is (a).

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Q137JEE Main 2025 · 29 JAN · SHIFT 1Medium

Relative Velocity in Two Dimensions · River & boat

The maximum speed of a boat in still water is 27 km/h. Now this boat is moving downstream in a river flowing at 9 km/h. A man in the boat throws a ball vertically upwards with speed of 10 m/s. Range of the ball as observed by an observer at rest on the bank is __________ cm. (Take $g = 10$ m/s2)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 2000

Boat speed relative to bank (downstream) $= 27 + 9 = 36\text{ km/h} = 10\text{ m/s}$ ; ball shares it.
Time of flight $T = \frac{2u}{g} = \frac{2(10)}{10} = 2\text{ s}$
Range seen from bank $= 10 \times 2 = 20\text{ m} = 2000\text{ cm}$

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Q138MHT CET 2025 · 25 APR · SHIFT 2Medium

Relative Velocity in Two Dimensions · Chasing & minimum distance

Two girls are standing at the ends ' $A$ ' and ' $B$ ' of a ground where $AB = b$ . The girl at ' B ' starts running in a direction perpendicular to ' $AB$ ' with velocity ' $V_1$ '. The girl at ' A ' starts running simultaneously with velocity ' $V_2$ ' and in shortest distance meets the other girl in time ' $t$ '. The value of ' $t$ ' is

Show answer and solution

Answer: (d)

Meeting point C: $BC = V_1 t$ , $AC = V_2 t$ , right angle at B.
$(V_2 t)^2 = b^2 + (V_1 t)^2$
$t = \frac{b}{\sqrt{V_2^2 - V_1^2}}$
So the answer is (d).

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Q139MHT CET 2025 · 25 APR · SHIFT 1Medium

Relative Velocity in Two Dimensions · Chasing & minimum distance

Two boys are standing at points A and B on ground, where distance $AB = x$ . The boy at B stars running perpendicular to $AB$ with velocity $v_1$ . The boy at A starts running simultaneously with velocity v and meets the other boy in time $t$ . The value of $t$ is

Show answer and solution

Answer: (c)

Meeting point C: $BC = v_1 t$ , $AC = vt$ , right angle at B.
$(vt)^2 = x^2 + (v_1 t)^2$
$t = \left[\frac{x^2}{v^2 - v_1^2}\right]^{1/2}$
So the answer is (c).

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Q140JEE Main 2023 · 31 JAN · SHIFT 1Easy

Relative Velocity in Two Dimensions · River & boat

The speed of a swimmer is $4\text{ km h}^{-1}$ in still water. If the swimmer makes his strokes normal to the flow of river of width $1\text{ km}$ , he reaches a point $750\text{ m}$ down the stream on the opposite bank. The speed of the river water is ___________ $\text{km h}^{-1}$

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 3

Strokes normal to flow: $t = \frac{d}{v_s} = \frac{1}{4}\text{ h}$
Drift $= v_r t \Rightarrow 0.75 = v_r \times \frac{1}{4}$
$v_r = 3\text{ km h}^{-1}$

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Q141JEE Main 2022 · 27 JUN · SHIFT 1Medium

Relative Velocity in Two Dimensions · Rain & umbrella

A girl standing on road holds her umbrella at 45 $^\circ$ with the vertical to keep the rain away. If she starts running without umbrella with a speed of 15 $\sqrt{2}$ kmh ${}^{-1}$ , the rain drops hit her head vertically. The speed of rain drops with respect to the moving girl is:

Show answer and solution

Answer: (c)

Umbrella at $45^\circ$ : rain's horizontal and vertical components are equal, $v_x = v_y$ .
Running at $15\sqrt{2}$ makes rain vertical, so $v_x = 15\sqrt{2} \Rightarrow v_y = 15\sqrt{2}\text{ km/h}$ .
Relative to the girl only $v_y$ remains: $15\sqrt{2} = \frac{30}{\sqrt{2}}\text{ km h}^{-1}$
So the answer is (c).

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Q142JEE Main 2022 · 26 JUN · SHIFT 1Medium

Relative Velocity in Two Dimensions · Seen from a moving frame

A fighter jet is flying horizontally at a certain altitude with a speed of 200 ms ${}^{-1}$ . When it passes directly overhead an anti-aircraft gun, a bullet is fired from the gun, at an angle $\theta$ with the horizontal, to hit the jet. If the bullet speed is 400 m/s, the value of $\theta$ will be ___________ $^\circ$ .

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 60

To hit the jet, the bullet's horizontal velocity must equal the jet's.
$400\cos\theta = 200 \Rightarrow \cos\theta = \frac{1}{2}$
$\theta = 60^\circ$

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Q143JEE Main 2021 · 16 MAR · SHIFT 2Easy

Relative Velocity in Two Dimensions · River & boat

A swimmer can swim with velocity of 12 km/h in still water. Water flowing in a river has velocity 6 km/h. The direction with respect to the direction of flow of river water he should swim in order to reach the point on the other bank just opposite to his starting point is ____________ $^\circ$ . (Round off to the Nearest Integer) (Find the angle in degrees)

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 120

To reach the opposite point, the component along the flow must cancel: $12\cos\theta + 6 = 0$
$\cos\theta = -\frac{1}{2} \Rightarrow \theta = 120^\circ$
Answer: $120^\circ$

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Q144JEE Main 2021 · 26 AUG · SHIFT 2Easy

Relative Velocity in Two Dimensions · Seen from a moving frame

A bomb is dropped by fighter plane flying horizontally. To an observer sitting in the plane, the trajectory of the bomb is a:

Show answer and solution

Answer: (c)

The bomb keeps the plane's horizontal velocity, so relative to the plane its horizontal velocity is zero.
Relative to the plane it only falls with acceleration $g$ .
So the answer is (c) a straight line vertically down.

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Q145JEE Main 2021 · 18 MAR · SHIFT 1Medium

Relative Velocity in Two Dimensions · River & boat

A person is swimming with a speed of 10 m/s at an angle of 120 $^\circ$ with the flow and reaches to a point directly opposite on the other side of the river. The speed of the flow is 'x' m/s. The value of 'x' to the nearest integer is __________.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 5

To reach directly opposite, the component along the flow must cancel: $x = 10|\cos 120^\circ|$
$x = 10 \times 0.5 = 5$
Answer: $5$

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Q146JEE Main 2021 · 20 JUL · SHIFT 1Medium

Relative Velocity in Two Dimensions · Seen from a moving frame

A butterfly is flying with a velocity $4\sqrt{2}$ m/s in North-East direction. Wind is slowly blowing at 1 m/s from North to South. The resultant displacement of the butterfly in 3 seconds is:

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Answer: (d)

Butterfly: $4\sqrt{2}$ at NE $= 4\hat{i} + 4\hat{j}$ (east, north). Wind: $-1\hat{j}$ .
Resultant $= 4\hat{i} + 3\hat{j}$ , speed $= 5\text{ m/s}$
Displacement in $3\text{ s} = 15\text{ m}$
So the answer is (d).

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Q147JEE Main 2021 · 27 JUL · SHIFT 2Difficult

Relative Velocity in Two Dimensions · River & boat

A swimmer wants to cross a river from point A to point B. Line AB makes an angle of 30 $^\circ$ with the flow of river. Magnitude of velocity of the swimmer is same as that of the river. The angle $\theta$ with the line AB should be _________ $^\circ$ , so that the swimmer reaches point B.

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 30

Swimmer velocity makes angle $30^\circ + \theta$ with the flow; river velocity is along the flow, both of equal magnitude.
The sum of two equal vectors bisects the angle between them, so the resultant is at $\frac{30^\circ + \theta}{2}$ from the flow.
Resultant must be along AB: $\frac{30^\circ + \theta}{2} = 30^\circ$
$\theta = 30^\circ$

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Q148JEE Main 2020 · 6 SEP · SHIFT 2Medium

Relative Velocity in Two Dimensions · Rain & umbrella

When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed v, he sees that rain drops are coming at an angle 60° from the horizontal. On further increasing the speed of the car to (1 + $\beta$ )v, this angle changes to 45o. The value of $\beta$ is close to:

Show answer and solution

Answer: (b)

Rain falls vertically with speed $u$ ; relative to the car it has horizontal component equal to car speed.
At $60^\circ$ from horizontal: $\tan 60^\circ = \frac{u}{v} \Rightarrow u = \sqrt{3}v$
At $45^\circ$ : $u = (1+\beta)v \Rightarrow 1 + \beta = \sqrt{3}$
$\beta = 0.732$ , so the answer is (b) 0.73.

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Q149NEET 2019Easy

Relative Velocity in Two Dimensions · River & boat

The speed of a swimmer in still water is 20 m/s. The speed of river water is 10 m/s and is flowing due east. If he is standing on the south bank and wishes to cross the river along the shortest path, the angle at which he should make his strokes w.r.t. north is given by:

Show answer and solution

Answer: (d)

For the shortest path the resultant must point due north, so he heads upstream (west of north).
$\sin\theta = \frac{v_r}{v_s} = \frac{10}{20} = \frac{1}{2} \Rightarrow \theta = 30^\circ$
So the answer is (d) $30^\circ$ west of north.

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Q150JEE Main 2019 · 9 APR · SHIFT 1Easy

Relative Velocity in Two Dimensions · River & boat

The stream of a river is flowing with a speed of 2km/h. A swimmer can swim at a speed of 4km/h. What should be the direction of the swimmer with respect to the flow of the river to cross the river straight?

Show answer and solution

Answer: (b)

To cross straight, the upstream component must cancel the flow: $v_s\cos\theta = -v_r$ , with $\theta$ measured from the flow.
$\cos\theta = -\frac{2}{4} = -\frac{1}{2} \Rightarrow \theta = 120^\circ$
So the answer is (b) $120^\circ$ .

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Q151JEE Main 2019 · 12 JAN · SHIFT 1Medium

Relative Velocity in Two Dimensions · Seen from a moving frame

A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle 60o with ground level. But he finds the aeroplane right vertically above his position. If v is the speed of sound, speed of the plane is:

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Answer: (d)

Sound travels from the old position along a line at $60^\circ$ to the ground, distance $vt$ .
In the same time the plane moves horizontally to directly overhead: $v_p t = vt\cos 60^\circ$
$v_p = \frac{v}{2}$
So the answer is (d).

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Q152JEE Main 2019 · 8 APR · SHIFT 1Difficult

Relative Velocity in Two Dimensions · Chasing & minimum distance

Ship A is sailing towards north-east with velocity $\vec{v} = 30\hat{i} + 50\hat{j}$ km/hr where $\hat{i}$ points east and $\hat{j}$ , north. Ship B is at a distance of 80 km east and 150 km north of Ship A and is sailing towards west at 10 km/hr. A will be at minimum distance from B in:

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Answer: (c)

Position of B relative to A: $\vec{r} = 80\hat{i} + 150\hat{j}$ ; $\vec{v}_{BA} = -10\hat{i} - (30\hat{i} + 50\hat{j}) = -40\hat{i} - 50\hat{j}$
Minimum distance at $t = -\frac{\vec{r}\cdot\vec{v}_{BA}}{|\vec{v}_{BA}|^2} = \frac{3200 + 7500}{1600 + 2500}$
$t = \frac{10700}{4100} \approx 2.6\text{ h}$
So the answer is (c).

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Q153JEE Main 2018 · 15 APR · SHIFT 2Difficult

Relative Velocity in Two Dimensions · Chasing & minimum distance

A man in a car at location Q on a straight highway is moving with speed $v$ . He decides to reach a point P in a field at a distance d from the highway (point M) as shown in the figure. Speed of the car in the field is half to that on the highway. What should be the distance RM, so that the time taken to reach P is minimum?

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Answer: (d)

Let $RM = x$ . Time $T = \frac{QM - x}{v} + \frac{\sqrt{x^2 + d^2}}{v/2}$
$\frac{dT}{dx} = -\frac{1}{v} + \frac{2x}{v\sqrt{x^2 + d^2}} = 0 \Rightarrow 2x = \sqrt{x^2 + d^2}$
$4x^2 = x^2 + d^2 \Rightarrow x = \frac{d}{\sqrt{3}}$
So the answer is (d).

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Q154NEET 2015 · AIPMT · CANCELLED PAPERMedium

Relative Velocity in Two Dimensions · Chasing & minimum distance

A ship A is moving Westwards with a speed of 10 km h−1 and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h ${}^{-1}$ . The time after which the distance between them becomes shortest, is

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Answer: (d)

Take A at origin. Relative to A, B starts at $(0, -100)$ with velocity $\vec{v}_B - \vec{v}_A = (10, 10)\text{ km/h}$ .
Closest approach when $\vec{r}\cdot\vec{v}_{rel} = 0$ : $10(10t) + 10(-100 + 10t) = 0$
$200t = 1000 \Rightarrow t = 5\text{ h}$
So the answer is (d).

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Q155JEE Main 2013 · 25 APR · SHIFT 1Difficult

Relative Velocity in Two Dimensions · Seen from a moving frame

The maximum range of a bullet fired from a toy pistol mounted on a car at rest is $R_0 = 40$ m. What will be the acute angle of inclination of the pistol for maximum range when the car is moving in the direction of firing with uniform velocity $v = 20$ m/s on a horizontal surface? ( $g = 10$ m/s $^{2}$ )

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Answer: (b)

At rest: $R_0 = \dfrac{u^2}{g} = 40 \Rightarrow u = 20$ m/s.
Car moving: horizontal speed $u\cos\theta + v$ , flight time $\dfrac{2u\sin\theta}{g}$ , so $R = \dfrac{2u\sin\theta}{g}(u\cos\theta + v)$ .
$\dfrac{dR}{d\theta} = 0 \Rightarrow u\cos 2\theta + v\cos\theta = 0 \Rightarrow 20(2\cos^2\theta - 1) + 20\cos\theta = 0$
$2\cos^2\theta + \cos\theta - 1 = 0 \Rightarrow \cos\theta = \dfrac{1}{2}$
So the answer is (b) $60^\circ$ .

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Q156NEET 2005 · AIPMTMedium

Relative Velocity in Two Dimensions · Chasing & minimum distance

Two boys are standing at the ends A and B of a ground where AB = a. The boy at B starts running in a direction perpendicular to AB with velocity v1. The boy at A starts running simultaneously with velocity v and catches the other in a time t, where t is

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Answer: (d)

Meeting point C: $BC = v_1 t$ , $AC = vt$ , right angle at B.
$(vt)^2 = a^2 + (v_1 t)^2$
$t = \sqrt{\frac{a^2}{v^2 - v_1^2}}$
So the answer is (d).

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Q157JEE Main 2004 · AIEEEMedium

Relative Velocity in Two Dimensions · Chasing & minimum distance

A ball is thrown from a point with a speed ν0 at an angle of projection θ. From the same point and at the same instant person starts running with a constant speed $\frac{v_0}{2}$ to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection θ?

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Answer: (c)

The person catches the ball if he always stays under it: horizontal speeds must match.
$v_0\cos\theta = \frac{v_0}{2} \Rightarrow \cos\theta = \frac{1}{2}$
$\theta = 60^\circ$ , so the answer is (c) Yes, $60^\circ$ .

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Q158NEET 2000 · AIPMTEasy

Relative Velocity in Two Dimensions · River & boat

The width of river is 1 km. The velocity of boat is 5 km/hr. The boat covered the width of river along the shortest path in 15 min. Then the velocity of river stream is

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Answer: (a)

Shortest path: resultant velocity is straight across, $v_{res} = \frac{1\text{ km}}{0.25\text{ h}} = 4\text{ km/h}$ .
$v_r = \sqrt{v_b^2 - v_{res}^2} = \sqrt{25 - 16} = 3\text{ km/h}$
So the answer is (a) 3 km/hr.

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Q159NEET 1999 · AIPMTMedium

Relative Velocity in Two Dimensions · River & boat

A person swims in a river aiming to reach exactly opposite point on the bank of a river. His speed of swimming is $0.5\text{ m/s}$ at an angle $120^\circ$ with the direction of flow of water. The speed of water in stream is

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Answer: (c)

To reach the opposite point, the swimmer's component along the flow cancels the stream.
$v_r = 0.5 \times |\cos 120^\circ| = 0.5 \times 0.5 = 0.25\text{ m/s}$
So the answer is (c).

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Q160NEET 1998 · AIPMTEasy

Relative Velocity in Two Dimensions · River & boat

The speed of a boat is $5\text{ km/h}$ in still water. It crosses a river of width $1.0\text{ km}$ along the shortest possible path in $15\text{ min}$ . The velocity of the river water is (in $\text{km/h}$ )

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Answer: (c)

Shortest path: resultant velocity is straight across, $v_{res} = \frac{1}{0.25} = 4\text{ km/h}$ .
$v_r = \sqrt{5^2 - 4^2} = 3\text{ km/h}$
So the answer is (c) 3.

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Q161NEET 1994 · AIPMTEasy

Relative Velocity in Two Dimensions · River & boat

A boat is sent across a river with a velocity of $8\text{ km h}^{-1}$ . If the resultant velocity of boat is $10\text{ km h}^{-1}$ , then velocity of river is

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Answer: (b)

Boat velocity $8\text{ km/h}$ is perpendicular to the river flow.
$v_{res}^2 = v_b^2 + v_r^2 \Rightarrow v_r = \sqrt{10^2 - 8^2} = 6\text{ km/h}$
So the answer is (b).

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Q162NEET 1993 · AIPMTEasy

Relative Velocity in Two Dimensions · River & boat

A boat is sent across a river with a velocity of $8\text{ km h}^{-1}$ . If the resultant velocity of boat is $10\text{ km h}^{-1}$ , then velocity of river is

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Answer: (b)

Boat velocity $8\text{ km/h}$ is perpendicular to the river flow.
$v_r = \sqrt{10^2 - 8^2} = 6\text{ km/h}$
So the answer is (b).

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