Answer to Question #165183 in Physics for Jason

Question

A ball is dropped from a tower with a height of 55 m. The ball weighs 2 kg. What is the ball’s potential energy? What is the ball’s kinetic energy just before it hits the ground? What is the ball’s velocity just before it hits the ground?

Answer 1

(a) By the definition of the potential energy, we have:

$$P=mgh=2\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot55\ m=1078\ J.$$

(b) According to the law of conservation of energy, the ball’s potential energy at the height of 55 meters is converted into the ball's kinetic energy just before it hits the ground:

$$PE=KE=1078\ J.$$

(c) We can find the ball’s velocity just before it hits the ground from the law of conservation of energy:

$$PE=KE=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{\dfrac{2KE}{m}}=\sqrt{\dfrac{2\cdot1078\ J}{2\ kg}}=32.83\ \dfrac{m}{s}.$$