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?

Respuesta :

Answer: 784 J

Explanation:

The kinetic energy [tex]K[/tex] of the bowling ball is given by the following equation:

[tex]K=\frac{1}{2}mV^{2}[/tex]

Where:

[tex]m=2 kg[/tex] is the mass of the ball

[tex]V=28 \frac{m}{s}[/tex] is the velocity of the ball before hitting the ground

Then:

[tex]K=\frac{1}{2}2kg(28 \frac{m}{s})^{2}[/tex]

[tex]K=784 J[/tex] This is the kinetic energy of the bowling ball

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