Projectile motionEdexcel A-Level Physics: Revision notes

A projectile is an object moving freely under gravity alone, once launched. The key idea is that the horizontal and vertical motions are independent: the vertical motion is not affected by the horizontal motion, and vice versa.

• Horizontally: no force (air resistance ignored), so acceleration is zero and the velocity is constant
• Vertically: the only force is weight, so acceleration is g = 9.81 m s⁻² downwards

The two motions share only the time of flight, so time is the link between horizontal and vertical calculations. A ball dropped and a ball fired horizontally from the same height hit the ground at the same time.

Launch at speed u horizontally from height h. Vertically: initial vertical velocity is zero, so the time of fall is found from h = ½gt², giving t = √(2h/g). Horizontally: the range is x = ut.

At any time the vertical velocity is v_y = gt and the horizontal velocity is still u. The speed is v = √(u² + v_y²) and the angle below the horizontal is tan θ = v_y/u.

Worked example. A stone is thrown at 12 m s⁻¹ from a cliff 45 m high: t = √(90/9.81) = 3.03 s, range = 12 × 3.03 = 36 m, v_y = 29.7 m s⁻¹, speed = 32 m s⁻¹.

Launch at speed u at angle θ above the horizontal. Resolve the velocity: u_x = u cos θ (constant) and u_y = u sin θ (changes by −g each second).

• Time to the highest point: u sin θ / g (vertical velocity is zero there)
• Maximum height: H = (u sin θ)²/2g
• Time of flight to the same level: T = 2u sin θ / g
• Range on level ground: R = u cos θ × T

At the highest point the vertical velocity is zero but the horizontal velocity u cos θ remains, so the speed is not zero. The acceleration is g downwards throughout. The path is a parabola, symmetrical about the highest point on level ground.

Worked example. A ball kicked at 20 m s⁻¹ at 30°: u_y = 10 m s⁻¹, u_x = 17.3 m s⁻¹, T = 2.04 s, H = 5.1 m, R = 35 m.

Apply the equations of uniform acceleration separately: vertically use a = −g (if upwards is positive) with s_y, u_y, v_y and t; horizontally use a = 0, so s_x = u_x t.

Strategy: (1) choose a sign convention; (2) resolve the launch velocity; (3) write down known values for each direction; (4) find the time from the vertical motion; (5) use the time for the horizontal motion. For landing at a lower level than launch, s_y is negative.

In practice, air resistance opposes the velocity. It reduces both horizontal and vertical speeds, so the real path is shorter in range and lower in height than the ideal parabola, and the descent is steeper than the rise. The effect is small for dense, slow objects over short distances (a steel ball) and large for light objects or high speeds (a shuttlecock).

The independence idea still holds approximately, and the parabola is a good model whenever drag is small compared with weight.