Learn9702Paper 2 · AS Level Structured QuestionsKinematicsProjectile motion

Kinematics · Subtopic 2 of 2

Projectile Motion

A projectile moves horizontally at constant velocity while accelerating vertically at $g$ . The two motions are independent and share only the time of flight. Resolve the launch velocity into components, solve the vertical motion for the time, then use it for the horizontal distance. Air resistance shortens the range and makes the path lopsided.

Cambridge 9231 Paper 3Cambridge 9702 Paper 1Cambridge 9702 Paper 2Edexcel YPH11 WPH11AQA 7408 Paper 1≈ 40 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Explain why horizontal and vertical motion are independent.
Solve problems for horizontally launched projectiles.
Solve problems for projectiles launched at an angle: time of flight, range and maximum height.
Describe the effect of air resistance on a trajectory.

Pause and think

At the same moment, one bullet is fired horizontally from a gun and another is dropped from the same height. Ignoring air resistance, which hits the ground first?

Section 2 of 14

Core idea

Start hereDropped and fired bullets land together

Fire a bullet horizontally and drop another from the same height at the same instant: on flat ground, they hit at the same time. Horizontal motion doesn't affect vertical motion. That independence is the whole key to projectiles.

The big idea

Split motion: horizontal constant velocity; vertical acceleration $g$ down. Time links the two.

1. Independence

Projectile launched at 20 m/s and angle θ

launch angle θ (°) = 15

Range is greatest at

45^\circ

(about 40.8 m). Angles that add to

90^\circ

(like

30^\circ

and

60^\circ

) give the same range.

x=u\cos\theta\,t,\qquad y=u\sin\theta\,t-\tfrac12gt^2

$u$: launch speed.
$\theta$: launch angle above horizontal.

2. Launched at an angle

Dropped and thrown together hit the ground at the same time

At each instant the two balls are at the same height: horizontal motion does not affect vertical motion.

Worked example20 m s⁻¹ at 30°

Level ground.

Goal: Time of flight, range, maximum height.
Strategy: Components; vertical for time.

$u_x=17.3,\ u_y=10.0$
$20\cos30^\circ$ , $20\sin30^\circ$ .
$0=10t-4.905t^2\Rightarrow t=2.04\text{ s}$
Back to $y=0$ .
$17.3\times2.04=35.3\text{ m}$
Range.
$0=10^2-2(9.81)h\Rightarrow h=5.10\text{ m}$
$v_y=0$ at the top.

Answer

2.04

35.3

5.10

✓ CheckRange formula

\frac{u^2\sin2\theta}g=\frac{400(0.866)}{9.81}=35.3

✓.

3. Horizontal launch

Worked exampleOff a 20 m cliff at 15 m s⁻¹

Horizontal launch.

Goal: Time, distance, impact velocity.
Strategy: Vertical from rest.

$20=4.905t^2\Rightarrow t=2.02\text{ s}$
Vertical.
$15\times2.02=30.3\text{ m}$
Horizontal.
$v_y=9.81\times2.02=19.8$
Final vertical speed.
$\sqrt{15^2+19.8^2}=24.8\text{ m s}^{-1}\text{ at }52.9^\circ$
Recombine; below horizontal.

Answer

2.02

30.3

24.8

m s

^{-1}

52.9^\circ

below horizontal.

✓ CheckEnergy:

\frac12v^2=\frac12(225)+9.81(20)

gives

v=24.8

✓.

Edge caseAir resistance Reduces range and height; the path becomes asymmetric (steeper descent).

Common misconceptionZero velocity at the top Only the vertical component is zero; the horizontal

u\cos\theta

remains.

Summary

Components.
Time links directions.

Projectile: Object moving under gravity alone after launch.

HistoryGalileo's parabola. Before Galileo, people thought a cannonball flew in a straight line, then suddenly dropped. In

1638

he showed that combining steady horizontal motion with uniformly accelerated falling gives a parabola, and that

45^\circ

gives the longest range.

Fun factThe basketball "hang time". A great basketball dunker seems to hang in the air. But a vertical leap of

1

m means a time in the air of only about

0.9

s, and half of it is spent in the top quarter of the jump, where the player moves slowest. That's the "hang".

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