Question

A student threw a ball horizontally out a window 8.0m above the ground. It was caught by another student who was 10.0m horizontally away from a spot directly below the first student.What was the velocity of the ball

Answer 1

The horizontal distance traveled by a ball

$$d=v_0t$$

The time of a ball free falling

$$t=\sqrt{2h/g}$$

So, the initial velocity of a ball

$$v_0=\frac{d}{t}=\frac{d}{\sqrt{2h/g}}$$

$$=\frac{10.0\:\rm m}{\sqrt{2\times 8.0\:\rm m/(9.8\:\rm m/s^2)}}=7.83\:\rm m/s$$