SOLUTION: If a ball rebounds three-fifths as far as it falls, how far will it (vertically) travel before coming to rest if dropped 14 feet?

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Question 552039: If a ball rebounds three-fifths as far as it falls, how far will it (vertically) travel before coming to rest if dropped 14 feet?
Found 2 solutions by Theo, ikleyn:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website!
this is actually 2 infinite geometric progressions.
the first geometric progression models the downward movement of the ball.
the second geometric progression models the upward movement of the ball.
the formula for an infinite geometric progression is:
S = a/(1-r)
a is the first value.
r is the common ratio
in this problem the common ratio is equal to 1/2.
this is because each time the downward movement of the ball is 1/2 what it was before.
the downward movement of the ball is modeled by the formula:
S = 14/.5
the upward movement of the ball is modeled by the formula:
S = 7/.5
you add these formulas together to get the downard and upward movement of the ball.
you get S = 14/.5 + 7/.5 = 28 + 14 = 42
the ball will travel 42 vertical units before it comes to rest.
28 of those vertical units will be in the downward direction.
14 of those vertical units will be in the upward direction.
a progression of what happens to the ball was modeled through excel.
the results are shown below:
trip equals the number of times up and down vertical movements were executed.
1 is the first time.
2 is the second time.
etc.
down is the number of units traveled down each time.
up is the number of units traveled up each time.
sum 1 is the cumulative sum of the units traveled down to that time.
sum 2 is the cumulative sum of the units traveled up to that time.
sum 3 is the cumulative sum of the units traveled up and down to that time.

trip down up sum 1 sum 2 sum 3 1 14.0000 7.0000 14.0000 7.0000 21.0000 2 7.0000 3.5000 21.0000 10.5000 31.5000 3 3.5000 1.7500 24.5000 12.2500 36.7500 4 1.7500 0.8750 26.2500 13.1250 39.3750 5 0.8750 0.4375 27.1250 13.5625 40.6875 6 0.4375 0.2188 27.5625 13.7813 41.3438 7 0.2188 0.1094 27.7813 13.8906 41.6719 8 0.1094 0.0547 27.8906 13.9453 41.8359 9 0.0547 0.0273 27.9453 13.9727 41.9180 10 0.0273 0.0137 27.9727 13.9863 41.9590 11 0.0137 0.0068 27.9863 13.9932 41.9795 12 0.0068 0.0034 27.9932 13.9966 41.9897 13 0.0034 0.0017 27.9966 13.9983 41.9949 14 0.0017 0.0009 27.9983 13.9991 41.9974 15 0.0009 0.0004 27.9991 13.9996 41.9987 16 0.0004 0.0002 27.9996 13.9998 41.9994 17 0.0002 0.0001 27.9998 13.9999 41.9997 18 0.0001 0.0001 27.9999 13.9999 41.9998 19 0.0001 0.0000 27.9999 14.0000 41.9999 20 0.0000 0.0000 28.0000 14.0000 42.0000

you can see that after 20 units it has sort of stabilized at total down movement of 28 and total up movement of 14 for total down and up movement of 42.
this is because of rounding to only the fifth decimal place.
more detail would have shown that 28 and 14 were not quite achieved but were getting closer and closer.
the actual formula for n iterations would be:
Sn = a(1-r^n)/(1-r)
for a = 14, the total down movement would be equal to 27.9999733.
for a = 7, the total up movement would be equal to 13.99998665
the sum would be equal to 41.99995995.
very close to 42 but not quite there.

Answer by ikleyn(52879)   (Show Source):
You can put this solution on YOUR website!
.
If a ball rebounds three-fifths as far as it falls, how far will it (vertically) travel before coming to rest
if dropped 14 feet?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

        The solution in the post by @Theo is incorrect.
        I came to bring a correct solution.

First rebound is = = ft. The heights of rebounds that follow make geometric progression with the common ratio of r = 3/5. The infinite sum of all the rebound heights series is = = = = 21. This value, 21 ft, we should double, counting each fall-rebund, and add the initial height of 14 ft. So, the final answer for the total travel is 14 + 2*21 = 14 + 42 = 56 ft.

Solved correctly.