Question

A cyclist cycles down a hill. At the top of the hill of height 50m he reaches a speed of 50ms-1. Calculate his speed at the bottom of the hill.

Answer 1

The speed of a cyclist at the bottom of the hill

$$v_b=\sqrt{v_t^2+2gh}\\ =\sqrt{50^2+2*9.8*50}=59\:\rm m/s$$