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Energy can be looked at at the apex of the ball's bounce and is proportional to h. Therefore, the energy retained between the first two bounces = 1.3/1.6 = 81%

Therefore, it loses about 19% of its energy per bounce.

To lose 90% means to retain 10% of the energy

0.8125^x=.1

x=log(.1)/log(.8125)

round up

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Q: A ball is dropped from a height of 1.60 m and rebounds to a height of 1.30 m How many rebounds will the ball make before losing at least 90 percent of its energy?
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Why does a body exerts more force on ground when it is dropped from a certain height than when it at rest at the ground although in both the cases mass and acceleration is constant See discussion too?

Interesting question. But when the object is at rest the potential energy of the object is 0, on the surface that is. When it is on a height h it's potential energy increase and when it is dropped from that height all that potential energy gets converted to kinetic energy just before hitting the ground. This extra force comes from this kinetic energy.


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When the ball hits the ground, kinetic energy is turned into potential energy and stored momentarily as the ball compresses. As the ball rebounds, the potential energy is converted back to kinetic energy. Some energy is also lost in the form of heat and air waves (sound). This continues with each bounce and due to the loss of energy, the ball gradually stops bouncing.


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Related questions

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