Projectile motion // physics rev

01 The big idea

A projectile is anything thrown or launched that then moves only under gravity. The trick is to treat the horizontal and vertical motion separately.

Horizontal: no force acts sideways, so the horizontal velocity stays constant the whole time.
Vertical: gravity pulls down, so the vertical velocity changes by 9.8 m/s every second (a = -9.8 m/s² if up is positive).
Split the launch velocity u at angle θ above the horizontal: u_x = u cos θ and u_y = u sin θ.
Time links both. The time in the air is the same for the horizontal and vertical motion. Usually you find t from the vertical part, then use it in the horizontal part.

x = u_x tv_y = u_y + a ts_y = u_y t + ½ a t²v_y² = u_y² + 2 a s_ys_y = ½(u_y + v_y) t

At the top of the path, v_y = 0 (but v_x is not zero). For a launch from flat ground the path is symmetric: time up = time down.

02 Simulator

Launch speed: 20 m/s Angle: 45° Launch height: 0 m Show velocity vectors

Time of flight:

Max height:

Range:

t: 0.00 s

v_x:

v_y:

Try the same speed at 30°, 45° and 60° from the ground. Which goes furthest?

03 Worked examples

Example 1: horizontal launch off a cliff

A ball is kicked horizontally at 12 m/s off a cliff 20 m high. Find (a) the time to hit the ground, (b) how far from the base it lands, (c) its speed on impact.

Step 1: Split it up. Horizontal: u_x = 12 m/s (constant). Vertical: u_y = 0 (launched horizontally), a = 9.8 m/s² down, s_y = 20 m down. Step 2 (a): Vertical, s = u t + ½ a t² with u_y = 0 20 = ½ × 9.8 × t² t² = 40 / 9.8 = 4.08 t = 2.02 s Step 3 (b): Horizontal, x = u_x t = 12 × 2.02 = 24.2 m Step 4 (c): Vertical speed at impact: v_y = u_y + a t = 0 + 9.8 × 2.02 = 19.8 m/s Horizontal speed is still 12 m/s. v = √(12² + 19.8²) = 23.2 m/s Direction: tan θ = 19.8 / 12, so θ = 59° below the horizontal.

Example 2: angled launch from the ground

A ball is launched at 25 m/s at 40° above the horizontal from flat ground. Find the time of flight, maximum height and range.

Step 1: Components u_x = 25 cos 40° = 19.15 m/s u_y = 25 sin 40° = 16.07 m/s Step 2: Time to the top (v_y = 0) v = u + a t → 0 = 16.07 − 9.8 t → t = 1.64 s Step 3: Time of flight (symmetric path) = 2 × 1.64 = 3.28 s Step 4: Max height, v² = u² + 2 a s 0 = 16.07² − 2 × 9.8 × H H = 16.07² / 19.6 = 13.2 m Step 5: Range = u_x × T = 19.15 × 3.28 = 62.8 m

04 Practice questions

Use g = 9.8 m/s². Try each one on paper first, then reveal.