Question 1116991: A rubber ball gets dropped on a hard surface, taking a sequence of bounces that are 3/5 as high as the preceding one. If the ball is dropped from 10 feet, what is the total vertical distance it has traveled after it hits the surface the 5th time?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! The first bounce(s1) it covers 10+(10/(5/3)) = 16 feet
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The second bounce(s2), it is (10/(5/3))+(10/(25/9)) = 9.6 feet
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The third bounce(s3), it is (10/25/9)+(10/(125/27)) = 5.76 feet and so forth
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The distance covered by each bounce is 3/5 the distance covered by the previous bounce
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The total distance covered is the sum of the partial sums,
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total distance = s1 + s2 + s3 + ...... + sn
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Note that the partial sums si form a geometric series where the common ratio is 3/5
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si = s1 * (3/5)^(n-1), where i = 1, n
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S(5) = 16 * (1 - (3/5)^5) / (1 - (3/5)) = 16 * (2882/3125) * (5/2) = 36.8896 feet
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Note the sum of the first n terms in a geometric sequence is
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S(n) = s(1) * (1 - r^n) / (1 - r), where s(1) is the first term and r is the common ratio, r not = 1
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