Question

A rock is dropped from a height of 2.7 m. How fast is it going when it hits the ground?

Answer 1

We can find the velocity of the rock when it hits the ground from the law of conservation of energy:

$$PE=KE,$$ $$mgh=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{2gh}=\sqrt{2\cdot9.8\ \dfrac{m}{s^2}\cdot2.7\ m}=7.3\ \dfrac{m}{s}.$$