How far a skateboarder leaving a 0.5 m ledge at 4 m/s travels

Answer: horizontal range = 1.28 m (fall time 0.319 s, impact speed 5.08 m/s).

What we know

  • Launch height: 0.5 m
  • Horizontal launch speed: 4 m/s

Step by step

Given
h = 0.5 m, v = 4 m/s
Formula
R = v·√(2h/g)
Substitute
R = 4 × √(2 × 0.5 / 9.80665)
Result
R = 1.28 m

All results

Horizontal range1.28 m
Fall time0.319 s
Impact speed5.08 m/s

Open the horizontal projectile calculator with these numbers → and change any of them.

The physics

A projectile launched horizontally from a height has two independent motions: it keeps its horizontal speed unchanged while gravity accelerates it downward. The vertical drop alone sets the flight time — t = √(2h/g), exactly the free-fall time for that height — and the horizontal range is just that time multiplied by the launch speed: R = v·√(2h/g).