Question

A stone is thrown horizontally at 20 m/s from the top of a cliff 63 m high. How fast is it moving the instant before it hits the ground?

Answer 1

Given:

$v_0=20\:\rm m/s$

$H=63\:\rm m$

The law of conservation of energy says

$$\frac{mv_0^2}{2}+mgH=\frac{mv^2}{2}$$

$$v=\sqrt{v_0^2+2gH}$$

$$v=\sqrt{20^2+2*9.8*63}=40.4\:\rm m/s$$