SimpleBouncingBall

# SimpleBouncingBall - Let us assume that the ball is not...

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Bouncing Ball In this module we examine a bouncing ball. We assume that the ball bounces with a certain coefficient of restitution , e . The height of the drop compared to the height of the rebound determines this coefficient. If the two heights are h 0 and h 1 , respectively, then the e = h 1 h 0 . It represents the change in the vertical speed at the point of impact. Let us consider how long a ball will bounce based on the height of the drop and the coefficient of restitution. The time from the ball being dropped to impact is given by the following formula. gt 0 2 2 = h 0 or t 0 = 2 h 0 g Between the first and second bounces, the height the ball reaches is h 1 . The time between those two bounces is given by the following formula. t 1 = 2 2 h 1 g = 2 2 e 2 h 0 g = 2 e 2 h 0 g The time between the n th bounce and the ( n + 1) st bounce is given by the following formula. t n = 2 2 e 2 n h 0 g = 2 e n 2 h 0 g

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Let us change the problem slightly to make our calculations a little easier.
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Unformatted text preview: Let us assume that the ball is not dropped, but propelled upward to the first height h . This makes t = 2 2 h g . Thus the length of time that the ball bounces is given by the following. T = 2 e n 2 h g n = ! " = 2 2 h g # e n n = ! " = 2 2 h g # 1 1 \$ e Let us now make some calculations using our formula. Suppose that h = 3_ ft and that e = .8 . Then T = 4.3184 _ s . If h = 10_ ft and that e = .99 , then T = 157.686 _ s . If h = 10_ ft and that e = .9999 , then T = 15768.6 _ s . In the last case the ball bounces 4.38016 hours. If e = 1 , then the impact is perfectly elastic and the ball will bounce forever. This is an application of the geometric series. Below is a graph of a few bounces for e = 2 3 . The second graph is for e = 9 10 ....
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SimpleBouncingBall - Let us assume that the ball is not...

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