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?
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.

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