Answer to Question #124629 in Physics for jn

Question

A boulder rolls over a 50m high cliff. How fast was the boulder moving if it was found 180m on the ground? At what time did the boulder hit the ground? At what velocity was it approaching the ground?

Answer 1

The time it took to reach the bottom is

$$t=\sqrt\frac{2h}{g}=\sqrt\frac{2\cdot50}{9.8}=3.2\text{ s}.$$

It approached the end of the cliff with the horizontal velocity of

$$v_h=\frac{D}{t}=\frac{180}{3.2}=53\text{ m/s}.$$

By the end of its trip the vertical component of velocity was

$$v_v=\sqrt{2gh}=\sqrt{2\cdot9.8\cdot50}=31\text{ m/s}.$$

The resultant velocity at which the boulder hit the ground was

$$v=\sqrt{v_h^2+v_v^2}=\sqrt{53^2+31^2}=62\text{ m/s}.$$