Question

A heavy object is dropped from a vertical height of 8.0 m. What is it’s speed when it hits the ground?

Answer 1

We can find the speed of the object when it hits the ground from the law of conservation of energy:

$$PE=KE,$$ $$mgh=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{2gh}=\sqrt{2\cdot 9.8\ \dfrac{m}{s^2}\cdot 8\ m}=12.52\ \dfrac{m}{s^2}.$$

Answer:

$v=12.52\ \dfrac{m}{s^2}.$