Question 109830
You can work this problem at least two ways. One way is to just do a table analysis of
what happens ... as follows
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Ball falls 16 feet and hits the floor first time. 16 feet ... 1 time
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Next ball rebounds to 8 ft and falls back 8 ft where it hits floor second time. Total distance
traveled to this point is 16 + 8 + 8 = 32 ft and 2 hits
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Ball rebounds to 4 ft and falls back 4 ft where it hits 3rd time.  Total distance is 32 + 4 + 4
= 40 ft and 3 hits
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Ball rebounds to 2 ft and falls back 2 ft where it hits 4th time. Total distance now is
40 + 2 + 2 = 44 ft and 4 hits.
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Ball rebounds to 1 ft and falls back 1 ft where it hits 5th time. Total distance  is now
44 + 1 + 1 = 46 ft and 5 hits
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Ball rebounds to 1/2 ft and falls back 1/2 ft where it hits 6th time. Total distance up
to this point is 46 + 1/2 + 1/2 = 47 ft and 6 hits.
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Ball rebounds to 1/4 ft and falls back 1/4 ft to hit 7th time. Total distance now is
47 + 1/4 + 1/4 = 47.5 ft and 7 hits
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Ball rebounds to 1/8 ft (0.125 ft) and falls back 1/8 ft (again 0.125 ft) and hits for the
8th time. Total distance now is 47.5 + 0.125 + 0.125 = 47.75 ft and 8 hits.
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The ball rebounds to 1/16 ft (0.0625 ft) and falls back the same distance. The total distance
is now 47.75 + 0.0625 + 0.0625 = 47.875 ft and 9 hits.
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Finally the ball rebounds to 1/32 ft (0.03125 ft) and falls back the same amount, hitting 
the floor for the 10th and final time. Total distance = 47.875 + 0.03125 + 0.03125 = 47.9325 ft.
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You could also do this problem using a geometric progression.
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Suppose that instead of dropping the ball from a height of 16 ft to start the problem, that
we shot it up 16 feet from floor level. The first cycle would be up 16 ft and down 16 ft for
a total of 32 ft. The next cycle would be up 8 ft and down 8 ft for a total of 16 ft. Each 
cycle thereafter would be half of the preceding cycle. So in this progression the first
term is 32 ft, the next 16, the next 8, and so on.
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If you define a as the first term, r as the common ratio (in this problem r = 1/2), n
as the number of terms to be considered (10 for this problem) and L as the last term, the
equation for the sum (S) of the terms in a geometric progression is:
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{{{S = (a - r*L)/(1-r)}}}
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in which {{{L = a*r^(n-1)}}}
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In this problem:
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{{{L = 32*(1/2)^(10 - 1)= 32*(1/2)^9 = 32*(1/(2^9))= 32*(1/512) = 0.0625}}}
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Now that we have L, we can return to the sum equation and substitute appropriate values:
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{{{S = (a - r*L)/(1-r) = (32-((1/2)*0.0625))/(1-(1/2)) = (32-0.03125)/(1/2)= 63.9375}}}
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So this tells us that the total distance traveled is 63.9375 ft. But don't forget that
in order to get this into the form of a geometric progression we added 16 feet on the
front end so that every cycle would be complete. Therefore, we need to subtract that 16 feet
off because the ball did not start at ground level. It was dropped from the height of
16 feet. Subtracting 16 feet results in 63.9375 ft - 16 ft = 47.9375 ft for the total
distance traveled during the cycles. This agrees with the answer we got the first way
we did it ...
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Hope this helps you to understand the problem and how it can be solved.
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