Question

A rock is thrown off the cliff horizontally with the speed of v = 12 m/s. The cliff is h=47 meter high. Find the speed of the rock once it hits the water in m/s. Give your answer with one decimal place.

Answer 1

Given:

$v_0=12\:\rm m/s$

$h=47\:\rm m$

The law of conservation of energy says

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

Hence

$$v=\sqrt{v_0^2+2gh}=\sqrt{12^2+2*9.8*47}=32.6\:\rm m/s$$