Question

Suppose that a person throws a ball upward at 15 m/s while standing on the edge of a cliff so that the ball can fall to the base of the cliff 50 m below. How long does it take the ball to reach the base of the cliff?

Answer 1

The time is

$$t=v/g+\sqrt{2(h+v^2/(2g))/g}=5.1\text{ s}.$$