Asked by Anonymous
A certain ball, when dropped from a height rebounds to one half of the original height.Suppose thos ball is dropped from a point 8ft above the ground, how far has it traveled, counting up and down distance only, when it hits the ground for the eight time?
Answers
Answered by
Reiny
distance of 1st bounce --- 8
distance of 2nd bounce --- 2(4) = 8
distance of 3rd bounce --- 2(2) = 4
distance of 4th bounce --- 2(1) = 2
distance of 5th bounce --- 2(1/2) = 1
distance of 6th bounce --- 2(1/4) = 1/2
distance of 7th bounce --- 2(1/8) = 1/4
distance of 8th bounce --- 2(1/16)= 1/8
Since there are only 8 terms, we can just add them up
I got a sum of 191/8
If there had been more terms, notice that , not counting the first term, we have a GP with a = 8 and r = 1/2
sum = 8 + sum(7)
= 8 + 8(1 - (1/2)^7)/(1- 1/2)
= 8 + 8(127/128) / (1/2)
= 8 + 8(127/128)(2)
= 191/8 , just as above
distance of 2nd bounce --- 2(4) = 8
distance of 3rd bounce --- 2(2) = 4
distance of 4th bounce --- 2(1) = 2
distance of 5th bounce --- 2(1/2) = 1
distance of 6th bounce --- 2(1/4) = 1/2
distance of 7th bounce --- 2(1/8) = 1/4
distance of 8th bounce --- 2(1/16)= 1/8
Since there are only 8 terms, we can just add them up
I got a sum of 191/8
If there had been more terms, notice that , not counting the first term, we have a GP with a = 8 and r = 1/2
sum = 8 + sum(7)
= 8 + 8(1 - (1/2)^7)/(1- 1/2)
= 8 + 8(127/128) / (1/2)
= 8 + 8(127/128)(2)
= 191/8 , just as above
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