Question

A 1.67kg rock is dropped from the top of a cliff that is 32.7m above the surface of a lake. What is the speed of the rock just before it hits the surface of the water

Answer 1

The law of conservation of energy says

$$E_i=E_f$$

Hence

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

$$v=\sqrt{2gh}=\sqrt{2\times 9.8\times 32.7}=25.3\:\rm m/s$$