Horizontal projectile from a cliff
KinematicsA ball rolls off a 20 m high cliff with a horizontal speed of 12 m/s. How long is it in the air, and how far from the base does it land? (Use g = 9.8 m/s², ignore air resistance.)
- Split the motion. Horizontal and vertical motion are independent. Vertically the ball starts with vy0 = 0 and accelerates at g; horizontally it moves at a constant 12 m/s.
- Time from the vertical drop. h = ½gt² → t = √(2h/g) = √(2 × 20 / 9.8) = √4.08 ≈ 2.02 s.
- Range from the horizontal motion. x = vxt = 12 m/s × 2.02 s ≈ 24.2 m.
- Unit check. √(m ÷ m/s²) = s ✓ and m/s × s = m ✓.
Time of flight ≈ 2.0 s; landing distance ≈ 24 m from the base of the cliff.
Common mistake the solver flags: using the 12 m/s in the vertical equation. The launch is horizontal, so the initial vertical velocity is zero.