Bouncing Ball


Hi, I am a sophomore taking general physics one and was given the following problem: "A ball falls off of a table and lands on the floor below. It then bounces up .5 of the distance it fell and repeats this process until it has bounced a total of three times. If its last bounce took it to a height of 1 m, what was the ball's total time in the air from its fall off the table to the rise after its last bounce?" My question is, "Can I assume that the initial velocity for each bounce is equal to half of the final velocity from the proceeding impact, in the opposite direction?"


To answer your question:
When a ball bounces (neglecting air resistance) it will rise to a height where its potential energy equals the kinetic energy with which it left the floor on the upward flight. The potential energy is given by m*g*h where m is the mass of the ball, g is the acceleration due to gravity and h is the height of the bounce. The * symbol denotes multiplication.

In the situation you describe where the height of the ball is reduced by half with each bounce, the energy of each bounce is half the energy of the previous bounce, since the potential energy is directly proportional to the height of the bounce, m and g being constant. This means that the kinetic energy which is given by 1/2*m*v2 is reduced by half. If we look at the kinetic energy in one bounce compared to the kinetic energy of the next bounce,
rather than
as you suggest.

I would look at the problem this way:
A 1 meter height after the third bounce implies a 2 meter height after the second bounce, a 4 meter height after the first bounce and a 8 meter initial height. Use the relationship between distance fallen and time of fall to calculate the time of the initial 8 meter drop. Then, since a ball on the way up is subject to exactly the same acceleration as a ball on its way down, recognize that the rise time for each bounce will equal the fall time. So calculate the fall time for a 4 meter fall and double it, the fall time for a 2 meter fall and double that, and finally the time for a 1 meter fall not doubled since we stop the problem with the ball at the 1 meter height. Adding all these times together should give you the answer.

Hope this helps.

J. D. Jones
M. Casco Associates