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05-25-2003, 03:54 AM | #31 | |
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05-25-2003, 03:59 AM | #32 |
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Actually, you don't need any calculus for the problem. You simply have to remember the vertical projectile motion equation (which you can rework to get time as a function of bounce height) and the fact that potential energy is mgh (so that if energy halves, height halves). This leads to a power series, whose sum is easy to calculate (if you don't remember the relation, you can find it online fairly quickly).
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05-25-2003, 04:13 AM | #33 |
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Yeah, the power series summation was the calculus I didn't remember (I didn't even remember the difference between a geometric series and a power series until I looked it up), but I assume it wouldn't be too hard to look up and figure out.
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05-25-2003, 04:28 AM | #34 |
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balls
May I point out that answers you obtain are not going to be exact. As the engineer said to the physcist ,close enough is good enough.
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05-25-2003, 04:54 AM | #35 | |
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Re: balls
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05-25-2003, 05:37 AM | #36 |
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balls2
I am not trying to be obtuse or nitpicking. but could you explain the conditions for an exact answer.
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05-25-2003, 06:00 AM | #37 | |
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Re: balls2
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edit: OK, I just worked this out, the exact solution would be: (1/5)^1/2 + [(4/5)^1/2]*[(2^1/2)/(2 - 2^1/2)] |
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05-25-2003, 06:27 PM | #38 |
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Or slightly more succinctly as:
(3 + 8^½) / 5^½ |
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