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Old 05-25-2003, 03:54 AM   #31
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Quote:
Originally posted by SULPHUR
Why is there finite time? Where is finite time first mentioned?
I haven't worked through the problem myself (my calculus is a little rusty), but people on this thread have said that when you sum up the time for each bounce, you get a total time of 2.61 seconds. From comments on the thread, it seems the time for each bounce forms a power series, and some power series have finite sums--see this page for a summary of the conditions when that's true.
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Old 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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Old 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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Old 05-25-2003, 04:28 AM   #34
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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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Old 05-25-2003, 04:54 AM   #35
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Quote:
Originally posted by SULPHUR
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.
For the problem as stated, you can get an exact answer. But the problem makes a lot of assumptions which aren't going to be true of real bouncing balls.
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Old 05-25-2003, 05:37 AM   #36
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I am not trying to be obtuse or nitpicking. but could you explain the conditions for an exact answer.
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Old 05-25-2003, 06:00 AM   #37
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Quote:
Originally posted by SULPHUR
I am not trying to be obtuse or nitpicking. but could you explain the conditions for an exact answer.
The ones mentioned in the original post--dropped from rest at a height of exactly 1 meter, G is exactly 10 meters/second^2, no friction, ball loses exactly half its energy on each successive bounce, time spent in contact with ground is vanishingly small.

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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Old 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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