Bounce Time for a Bouncing Ball

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Click "toss the ball" for an animation. Initially, the ball is tossed horizontally with speed 1 m/s from a height of 1.5 m onto a flat floor. The ball has normal and tangential coefficients of restitution of 0.7 and 0.8. (That means the normal and tangential speeds are reduced by factors of 0.7 and 0.8 at each bounce.) It takes 3.10 seconds for the ball to bounce 13 times, and then it has moved horizontally 1.96 m. The formulas below show the ball has finished bouncing at time 3.135, and it is stopped at (1.96, 0).
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Contributed by: Roger B. Kirchner (February 2010)
Based on programs by: Enrique Zeleny, Rob Morris, and Oleksandr Pavlyk
Open content licensed under CC BY-NC-SA
Snapshots
Details
The program for computing the bouncing ball's path is adapted from the bouncing ball example in the Mathematica tutorial "EventLocator" Method for NDSolve, and Enrique Zeleny's Demonstration Stroboscopic Photograph of a Bouncing Ball.
The program for animating the ball is adapted from Rob Morris and Oleksandr Pavlyk's Demonstration Animated Projectile Motion.
The bouncing ball program assumes that if the tangential and normal components of velocity are and
before a bounce, they are
and
after the bounce, where
. We have generalized so the reflected components are
and -
, where we call
the tangential coefficient of restitution and
the normal coefficient of restitution. There may be a reason to assume
, but
is the usual coefficient of restitution, and
seems to be a frictional effect. Of course, friction would affect rotation, which we ignore.
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