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 Bouncy Bouncy part 2 (Posted on 2006-02-10)
Part 2: Super-ball

A small ball is thrown toward an incline. This (amazing) ball bounces perfectly.
It is thrown to the right horizontally from the point (0,5) and follows the path of the parabolic equation y=5 - (x^2)/18.
The incline follows the equation y=x/2.

Find the equation of the path the ball takes after its first bounce off of the incline.

(If you wish, find some of the subsequent bounces.)
(Assume that gravity pulls straight down on the ball. Also assume the ball is not spinning so that it is perfectly reflected.)

 See The Solution Submitted by Jer Rating: 4.5000 (2 votes)

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 re(3): solution -- barring any mistakes along the way | Comment 8 of 14 |
(In reply to re(2): solution -- barring any mistakes along the way by Hugo)

While my initial post expressed the thought that kinetic energy vs potential energy would be the key to solving the puzzle, in the actual solution I bypassed that by using a generic formula for the height,when thrown upward (or the upward component), y = a t - 16 t^2   (as you'll see the 16 doesn't matter). Then y' = a - 32 t, which is zero at maximum height when t = a/32, when the height y = a^2/32 - 16 a^2 / 32^2. But the latter is just k a^2 for some k or other. Then the solution just depended on the fact that the ratio of the vertical velocities at initial height y=3 was 1.8 (based on the different angle but the same speed).  Regardless of the value of k or the gravitational constant, the height above y=3 that would be achieved would be on the ratio of 1.8^2 to 1.
 Posted by Charlie on 2006-02-12 10:14:53

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