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

Step-by-step explanation:

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

`K=(1)/(2)mV^(2)`

Where:

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

`V=28 (m)/(s)` is the velocity of the ball before hitting the ground

Then:

`K=(1)/(2)2kg(28 (m)/(s))^(2)`

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