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Topic: The Bouncy Math of Willy Wu. (Read 361 times) |
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rloginunix
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The Bouncy Math of Willy Wu.
« on: Dec 11th, 2014, 9:05am » |
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The Bouncy Math of Willy Wu.
Willy Wu climbed to an initial height of H0 and launched strictly horizontally a small solid ball (which is a perfect sphere of uniform density) with an initial velocity of V0.
The small ball bounces off of a hard perfectly horizontal surface in such a way that the ratio of its vertical velocity after the bounce to its vertical velocity before the bounce is constant, known and given, n. In other words the vertical velocity of the ball diminishes at a constant rate.
There is no friction between the ball and the surface and, consequently, the horizontal velocity of the ball remains the same. There is no air drag, the small ball can be considered a material point.
Willy Wu wants to know - how much horizontal distance will the ball cover by the time it stops bouncing?
Willy Wu Tang gives an extra credit to those who are willing to analyze the case when the object is launched under a given angle  to the horizon.
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towr
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Re: The Bouncy Math of Willy Wu.
« Reply #1 on: Dec 11th, 2014, 10:39am » |
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I'd start by throwing the ball up to a height H0. This adds half a "bounce", but now we only have to count complete "bounces"
The time to reach the starting height is  (2 * H0 / g ), and thus requires a vertical velocity Vv of (2 * H0 * g )
The time for half a bounce from floor to top of the arc is Vv / g * nk for the kth bounce (starting at k=0 at t= - (2 * H0 / g ) )
times 2 for a whole bounce, summed over k = 0 to infinity
So the ball will be in the air for a time of 2 * Vv / g *   k=0 nk = 2 * Vv / g / (1-n) [edit]removed errant factor n[/edit]
But we need to subtract that initial time I added for throwing it up to the starting height H0,
so this gives 2 * Vv / g / (1-n) - (2 * H0 / g) to cover horizontal distance from t = 0
Multiply it by the initial horizontal velocity Vh = V0 and we're done.
I suppose a prettier answer would be gotten by asking for the distance covered from the first bounce on.
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| « Last Edit: Dec 11th, 2014, 10:25pm by towr » |
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rloginunix
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Re: The Bouncy Math of Willy Wu.
« Reply #2 on: Dec 11th, 2014, 1:16pm » |
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towr bounced right back (as usual).
Well done.
I started counting the full bounces past the first one, so I added half a bounce to the sum which starts with the index k = 1 but in the end it does not matter.
The answer is a pretty one. You do the honors since you did all the hard work.
When you sum the bounces starting from the k index of 0 the infinite series sums to 1/(1 - n). It sums to n/(1 - n) if you start counting the bounces from the index k = 1 (and then add half a bounce). Substitute V0 for Vy ...
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towr
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Re: The Bouncy Math of Willy Wu.
« Reply #3 on: Dec 11th, 2014, 10:42pm » |
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Ah, duh. And here I was think if I can just get rid of that ugly last term, when in fact I can just replace it by Vv/g (or do the reverse replacement)
So, (2 * H0 / g) * V0 * (1+n)/(1-n)
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| « Last Edit: Dec 11th, 2014, 10:43pm by towr » |
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rloginunix
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Re: The Bouncy Math of Willy Wu.
« Reply #4 on: Dec 12th, 2014, 7:39am » |
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Yep. You got it.
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