Question

Directions: Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground.
As it is just about to hit the ground from a fall off a building that is 40 meters tall and travelling 28 meters per second

What is the kinetic energy of the ball just before it hits the ground?

Answer 1

Answer: 784 J

Explanation:

The kinetic energy $K$ of the bowling ball is given by the following equation:

$K=\frac{1}{2}mV^{2}$

Where:

$m=2 kg$ is the mass of the ball

$V=28 \frac{m}{s}$ is the velocity of the ball before hitting the ground

Then:

$K=\frac{1}{2}2kg(28 \frac{m}{s})^{2}$

$K=784 J$ This is the kinetic energy of the bowling ball