Question

A cannonball is dropped from the top of a building. If the point of release is 105 ft above the ground, what is the velocity of the cannonball just before it strikes the ground?

Answer 1

The law of conservation of energy gives

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

So, the velocity of the cannonball just before it strikes the ground

$$v=\sqrt{2gh}$$

$$=\sqrt{2\times 32\:\rm ft/s^2\times 105\: ft}=82\:\rm ft/s=25\:\rm m/s$$