Projectile motion

A projectile is anything thrown or launched that then moves under gravity alone. The trick is to split the motion into two independent parts: horizontally the velocity stays constant, while vertically the object accelerates downwards at 9.8 m/s², just like a dropped ball. Vertical projectiles (thrown straight up) use only the vertical part.

Key facts

  • Ignoring air resistance, horizontal velocity is constant and the acceleration is 9.8 m/s² downwards.
  • Horizontal and vertical motions are independent; they share only the time.
  • Launch at speed u, angle θ: uₓ = u cos θ, u_y = u sin θ.
  • At the top of the path the vertical velocity is zero.
  • On level ground: time of flight = 2u sin θ / g; range = u² sin 2θ / g (greatest at 45°).
  • Speed at any moment = √(vₓ² + v_y²).

Worked examples

Example 1

A ball rolls off a 1.25 m high table at 3.0 m/s. How far from the table does it land? (g = 9.8 m/s²)

  1. Vertically it starts from rest and falls 1.25 m, just like a dropped ball. 1.25 = ½ × 9.8 × t², t = 0.505 s
  2. Horizontally it keeps going at 3.0 m/s the whole time. x = 3.0 × 0.505 = 1.52 m

Answer: 1.52 m

Example 2

A ball is thrown vertically upwards at 12 m/s. Find its maximum height and the time to get there. (g = 9.8 m/s²)

  1. At the highest point it stops for an instant: v = 0.
  2. Use v² = u² − 2gs. s = 12² ÷ 19.6 = 7.35 m
  3. Use v = u − gt. t = 12 ÷ 9.8 = 1.22 s

Answer: 7.35 m; 1.22 s

Example 3

A golf ball leaves the ground at 30 m/s, 40° above the horizontal. Find its time of flight, range and greatest height. (g = 9.8 m/s²)

  1. Split it into horizontal and vertical parts. uₓ = 30 cos 40° = 22.98 m/s, u_y = 30 sin 40° = 19.28 m/s
  2. Time of flight: up and back down, using the vertical part only. t = 2 × 19.28 ÷ 9.8 = 3.94 s
  3. Range: the horizontal speed never changes. x = 22.98 × 3.94 = 90.4 m
  4. Greatest height: vertical speed reaches 0. h = 19.28² ÷ 19.6 = 19 m

Answer: 3.94 s; 90.4 m; 19 m