Projectile Motion: theory | Physics Circle

CONCEPT

Motion in a Plane

In one line
In a plane, treat x and y as two separate straight-line motions joined by vectors: r gives position, its rate gives v, and the rate of v gives a.

POSITION

3.1.1 Position and displacement

Two coordinates change with time: the body moves in a plane.
Examples: a ball thrown at an angle, a car on a turn, a boat crossing a river.

r = x î + y ĵ

Δr = r₂ − r₁ = Δx î + Δy ĵ

Displacement joins P₁ to P₂; velocity is along the tangent

VELOCITY

3.1.2 Velocity and acceleration

v = drdt = vₓ î + v_y ĵ

v = √vₓ² + v_y² · tan θ = v_yvₓ

Components of velocity

a = dvdt = aₓ î + a_y ĵ

Instantaneous velocity is always along the tangent to the path.
r = a cos ωt î + a sin ωt ĵ: v ⟂ r and a = −ω²r (circle).

PATH

3.1.3 Equation of the path

Given x(t) and y(t): remove t to get y as a function of x.
x linear and y quadratic in t ⇒ parabola.
x = a sin ωt, y = a cos ωt ⇒ x² + y² = a², a circle.
Given v: use dydx = v_yvₓ and integrate.

CONSTANT a

3.1.4 Constant acceleration in a plane

Each component obeys the 1-D equations on its own.

v = u + at

r = ut + ½at²

Write x and y separately: x = uₓt + ½aₓt², y = u_yt + ½a_yt².
Find t from one direction, then use it in the other.

AVERAGE

3.1.5 Average velocity and acceleration

v_av = ΔrΔt · a_av = ΔvΔt

Only the end points count for average velocity.
Speed unchanged but direction turned ⇒ Δv ≠ 0.
Turning through 90° at speed v: |Δv| = v√2.

Practise PYQs on this topicMotion in a Plane (2-D Kinematics) · 28 questions with solutions

CONCEPT

Projectile Basics & Trajectory

In one line
A projectile moves under gravity alone: uniform motion sideways and free-fall motion up and down, which together trace a parabola.

PROJECTILE

3.2.1 What is a projectile?

A body in flight under the effect of gravity alone, with no fuel pushing it.
Examples: a bullet from a gun, an arrow, a javelin, a bomb released from a plane.

Three types of projectile motion

ASSUMPTIONS

3.2.2 Assumptions

No air resistance.
Curvature and rotation of the earth are ignored.
g is constant in magnitude and direction.

INDEPENDENCE

3.2.3 Independence of motions

Horizontal and vertical motions take place independently.
Horizontal: no force ⇒ uniform velocity u cos θ.
Vertical: acceleration g downward, starting with u sin θ.

Stays constant	Keeps changing
horizontal velocity u cos θ	vertical velocity
acceleration g	speed, velocity, momentum
mechanical energy	KE and PE

TRAJECTORY

3.2.4 Equation of trajectory

Velocity u at angle θ: components, H and R

x = (u cos θ) t

y = (u sin θ) t − ½gt²

Put t = xu cos θ in y:

y = x tan θ − gx²2u² cos²θ

Of the form y = ax − bx²: the path is a parabola.
Compare a given equation with it: a = tan θ, b gives u.

Same equation in terms of the range R:

y = x tan θ (1 − xR)

Position known at a time t: x gives u cos θ, y gives u sin θ.

ELEVATION

3.2.5 Elevation of the highest point

Top seen from the point of projection

tan φ = HR2 = 12 tan θ

Practise PYQs on this topicProjectile Basics & Trajectory · 10 questions with solutions

CONCEPT

Velocity, Momentum & Energy in Flight

In one line
The horizontal velocity never changes; the vertical velocity falls by g every second, so the top has only u cos θ.

VELOCITY

3.3.1 Velocity at any instant

vₓ = u cos θ · v_y = u sin θ − gt

v = √u² + g²t² − 2ugt sin θ

tan α = u sin θ − gtu cos θ = tan θ − gtu sec θ

Velocity is along the tangent: at the top only vₓ is left

Direction changes from θ above to θ below the horizontal.
Same horizontal velocity at two points: v₁ cos α₁ = v₂ cos α₂.
At the same level on the way down: vertical part reversed, v = u cos θ î − u sin θ ĵ.

TOP

3.3.2 At the highest point

v_y = 0, but the velocity is not zero.
Acceleration is still g, downward.
Velocity ⟂ acceleration only at the top.

v_top = u cos θ

Speed at the top is u ÷ 2 ⇒ θ = 60°.
Angle given with the vertical: change it to the horizontal first (90° − angle).

MOMENTUM

3.3.3 Momentum

Horizontal momentum mu cos θ stays constant.
Vertical momentum is zero at the top.

ENERGY

3.3.4 Energy in flight

Kinetic energy at projection K = ½mu².

K_top = ½mu² cos²θ = K cos²θ

PE_top = mgH = ½mu² sin²θ = K sin²θ

Sum = ½mu²: total energy is conserved.
Loss of KE from projection to top = ½m(u sin θ)².
PE ratio of two bodies at their tops = ratio of their heights.

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Corner

Change in velocity and momentum

Only the vertical part of velocity changes; the horizontal part stays u cos θ.
Projection point to top:

Δv = −u sin θ ĵ · Δp = −mu sin θ ĵ

Whole flight (back to the same level):

Δv = −2u sin θ ĵ · Δp = −2mu sin θ ĵ

Δp = mgT always points straight down.

Angular momentum about the point of projection

At the top, r ⟂ p with lever arm H.

L = mu cos θ · H = mu³ cos θ sin²θ2g

Direction: perpendicular to the plane of motion (into the page for a throw to the right).

Practise PYQs on this topicVelocity, Momentum & Energy in Flight · 16 questions with solutions

CONCEPT

Time of Flight, Maximum Height & Range

In one line
Time and height depend only on the vertical part u sin θ; the range multiplies the time of flight by the horizontal part u cos θ.

TIME

3.4.1 Time of flight

Time to go up: 0 = u sin θ − gt.
Time up = time down, so double it.

T = 2u sin θg = 2u_yg

Caught after T: time to the top = T2.

HEIGHT

3.4.2 Maximum height

H = u² sin²θ2g = u_y²2g

H ∝ u²: double the speed, four times the height.
Straight up (θ = 90°): H_max = u²2g.

Links between T and H:

H = gT²8 · T = 2√2Hg

RANGE

3.4.3 Horizontal range

R = u cos θ × T = u² sin 2θg = 2uₓu_yg

Maximum when sin 2θ = 1, i.e. θ = 45°.

R_max = u²g = 2H_max

At θ = 45°: H = R_max4.

R AND H

3.4.4 Relation between R and H

R = 4H cot θ

R = nH ⇒ tan θ = 4n.
R = H ⇒ tan θ = 4 (θ ≈ 76°).
R = 4H ⇒ θ = 45°.

SAME HEIGHT

3.4.5 Same height at two instants

Height y is crossed at t₁ (going up) and t₂ (coming down)

t₁ + t₂ = 2u sin θg · y = gt₁t₂2

Ball reaching a roof: use y = u sin θ t − ½gt² at the landing time.

AREA

3.4.6 Bullets fired in all directions

The farthest any bullet reaches is R_max: the wet or hit region is a circle.

A = π R_max² = πu⁴g²

A ∝ u⁴: double the speed, 16 times the area.
Same trajectory on another planet: g ∝ u².

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Corner

Two heights crossed twice

A, D at one level and B, C at a higher level h above them.
Time gaps t₂ (between A and D) and t₁ (between B and C):

t₂² − t₁² = 8hg

Practise PYQs on this topicTime of Flight, Maximum Height & Range · 34 questions with solutions

CONCEPT

Comparing Two Projectiles

In one line
Write each quantity as a power of u and a function of θ, then cancel what is common: same speed compares angles, same height or time compares u sin θ.

PROPORTION

3.5.1 What each quantity depends on

Quantity	Depends on	Same speed u
T	u sin θ	T ∝ sin θ
H	(u sin θ)²	H ∝ sin²θ
R	u² sin 2θ	R ∝ sin 2θ

Different speeds and angles: keep both, e.g. R₁ : R₂ = u₁² sin 2θ₁ : u₂² sin 2θ₂.
H ∝ T²: height 8 times ⇒ time 2√2 times.

COMPLEMENTARY

3.5.2 Complementary angles θ and 90° − θ

30° and 60° land at the same point; 45° goes farthest

sin 2(90° − θ) = sin 2θ: the ranges are equal.
Times, heights and paths are different.
Same is true for 45° − α and 45° + α.

T₁T₂ = tan θ · H₁H₂ = tan²θ

T₁T₂ = 2Rg

H₁ + H₂ = u²2g · R = 4√H₁H₂

Two times of flight T₁, T₂ for the same range: R = gT₁T₂ ÷ 2.
Angles 42° and 48°: same R; the larger angle goes higher.

SAME HEIGHT

3.5.3 Same height or same time to the top

Same H or same time to the top ⇒ same vertical velocity.

u₁ sin θ₁ = u₂ sin θ₂

u₁u₂ = sin θ₂sin θ₁

Same time of flight also means the same height.
One thrown straight up, other at an angle, same time in air: same height.

RATIOS

3.5.4 Quick ratios at the same speed

Angles	T₁ : T₂	H₁ : H₂	R₁ : R₂
30°, 60°	1 : √3	1 : 3	1 : 1
45°, 30°	√2 : 1	2 : 1	2 : √3
15°, 45°	—	—	1 : 2

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Corner

One projectile seen from another

Both have the same acceleration g, so the relative acceleration is zero.
Relative velocity stays constant: one sees the other move in a straight line.

y₁ − y₂x₁ − x₂ = u₁ sin θ₁ − u₂ sin θ₂u₁ cos θ₁ − u₂ cos θ₂ = constant

Angles 45° − α and 45° + α

Same range, equal to the range at 45° multiplied by cos 2α.

R = u² cos 2αg

Practise PYQs on this topicComparing Two Projectiles · 31 questions with solutions

CONCEPT

Horizontal Projectile & Inclined Plane

In one line
Thrown horizontally from a height: the sideways speed u stays, the fall is free fall from rest, so the time depends only on the height.

SET-UP

3.6.1 Thrown horizontally from height h

Initial velocity u is horizontal; initial vertical velocity is zero.
Take the downward direction as positive for y.

Ball thrown horizontally from a tower of height h

x = ut · y = ½gt²

y = gx²2u²

The path is a parabola starting horizontally.

TIME

3.6.2 Time of flight and range

T = √2hg

R = uT = u√2hg

T does not depend on u: a dropped body and a thrown body land together.
Mass, size and shape do not matter (no air resistance).

Dropped and thrown at different speeds: all at the same height at every instant

VELOCITY

3.6.3 Velocity at time t and on landing

v = √u² + g²t² · tan φ = gtu

φ is the angle below the horizontal.

v_ground = √u² + 2gh

AEROPLANE

3.6.4 Bomb or packet from an aeroplane

The bomb keeps the plane’s speed u: it is directly below the plane at every instant.
Release the bomb when the target is a horizontal distance R = u√(2h ÷ g) ahead.

Line-of-sight distance = √R² + h²

Displacement of the landing point from the release point = √(R² + h²).
Convert km/h to m/s: multiply by 5 ÷ 18.

VEHICLE

3.6.5 Dropped from a moving vehicle

The ball takes the vehicle’s velocity at the moment of release.
After release, its only acceleration is g, even if the vehicle was accelerating.
Bodies thrown towards each other horizontally from equal heights meet at the same height.

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Corner

Projectile on an inclined plane

Take axes along and perpendicular to the incline (angle α).
u_∥ = u cos θ, u_⊥ = u sin θ; a_∥ = −g sin α, a_⊥ = −g cos α.

θ is measured from the incline

T = 2u sin θg cos α · H = u² sin²θ2g cos α

R = 2u² sin θ cos(θ + α)g cos²α

Maximum range up the plane at θ = 45° − α ÷ 2:

R_max = u²g(1 + sin α) up · u²g(1 − sin α) down

Practise PYQs on this topicHorizontal Projectile & Inclined Plane · 12 questions with solutions

CONCEPT

Relative Velocity in Two Dimensions

In one line
Velocity of A as seen by B is v_A − v_B: add −v_B to v_A as vectors and read off the size and direction.

DEFINITION

3.7.1 Relative velocity

v_AB = v_A − v_B

v_AB = √v_A² + v_B² − 2v_Av_B cos θ

θ is the angle between v_A and v_B.
Work in components when directions are given as north, east, and so on.

RAIN

3.7.2 Rain and umbrella

Rain relative to the man: v_r − v_m

v_rm = v_r − v_m

tan θ = v_mv_r (rain falling vertically)

Hold the umbrella along v_rm, tilted forward by θ from the vertical.
Running makes rain look vertical ⇒ his speed equals the horizontal part of the rain velocity.

RIVER

3.7.3 River and boat

Boat speed v (in still water), river speed u, width d.

Head straight across for least time; head upstream for the shortest path

Least time: head straight across.

t_min = dv · drift = u dv

Shortest path (reach the opposite point, v > u): head upstream at θ from the normal.

sin θ = uv · t = d√v² − u²

Angle with the flow direction = 90° + θ, e.g. 120° when u = v ÷ 2.
Only the component across the river decides the crossing time.

CHASE

3.7.4 Chasing at right angles

B runs at V₁ perpendicular to AB = a; A runs at V and meets B.
A runs along the hypotenuse of a right triangle:

t = a√V² − V₁²

FRAME

3.7.5 Seen from a moving frame

Bomb from a plane: seen from the plane, a straight line vertically down.
Wind: add the wind velocity to the body’s velocity in still air.
Catching a ball while running at v: v = u cos θ.
Two ships: closest approach when r ⟂ relative velocity.

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Corner

Shortest distance between two moving bodies

Go to the frame of A: B moves with v_BA = v_B − v_A in a straight line.
Drop a perpendicular from A on this line: that is d_min.

B as seen by A

t = −r·vv² (r, v: B relative to A)

Firing from a moving vehicle

A gun on a car moving at v fires at speed u relative to the car.
Ground velocity of the bullet = (u cos θ + v) î + u sin θ ĵ; the time of flight is unchanged.

Practise PYQs on this topicRelative Velocity in Two Dimensions · 31 questions with solutions

EXAM-DAY CARD

Just see this before the exam

v = drdt

motion in a plane

y = x tan θ − gx²2u² cos²θ

trajectory

T = 2u sin θg

time of flight

R = u² sin 2θg

range

T = √2hg

horizontal

v_AB = v_A − v_B

relative

FORMULA SHEET

Whole chapter on one page

Plane

v = u + at

r = ut + ½at²

v_av = ΔrΔt

90° turn: |Δv| = v√2

Trajectory

y = x tan θ (1 − xR)

tan φ = ½ tan θ

v_top = u cos θ

KE_top = K cos²θ

T, H, R

T = 2u sin θg

H = u² sin²θ2g

R = u² sin 2θg

H = gT²8

Shortcuts

R = 4H cot θ

R_max = u²g = 2H_max

area = πu⁴g²

y = gt₁t₂2

Two projectiles

θ, 90° − θ: same R

T₁T₂ = 2Rg

R = 4√(H₁H₂)

same H: u₁ sin θ₁ = u₂ sin θ₂

Horizontal · River

T = √2hg, R = uT

v = √u² + 2gh

t_min = dv; sin θ = uv

rain: tan θ = v_mv_r

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Velocity is zero at the top

Angle given with the vertical used as θ

Same range means same height

A faster horizontal throw lands later

CORRECT ✓

v = u cos θ at the top; only v_y = 0

θ = 90° − angle with the vertical

θ and 90° − θ: same R, different H and T

T = √(2h ÷ g), whatever u is