We might think that a ball that is dropped from a height of 15 feet and rebounds
ID: 2872869 • Letter: W
Question
We might think that a ball that is dropped from a height of 15 feet and rebounds to a height 5/8 of its previous height at each bounce keeps bouncing forever since it takes infinitely many bounces. This is not true! We examine this idea in this problem. Show that a ball dropped from a height h feet reaches the floor in j /h seconds. Then use tins result to fmd the tune, in seconds, the ball has been bouncing when it hits the floor for the first, second, third and fourth times: time at first bounce = 1/4sqrt(15) time at second bounce = 1/4sqrt(15)+1/2sqrt(15*(5/8)) time at third bounce = 1/4sqrt(15)+1/2sqrt(15*(5/8))+1/2sqrt(15*(5/8)A2 time at fourth bounce = 1/4sqrt(15)+1/2sqrt(15*(5/8))+1/2sqrt(15*(5/8)A2 How long, in seconds, has the ball been bouncing when it hits the floor for the 71th time (find a closed form expression)? time at Tith bounce = What is the total time that the ball bounces for? total time=Explanation / Answer
B)nth bounce: (1/4)sqrt15 +(1/2)sqrt(15*(5/8))+(1/2)sqrt(15*(5/8)2)+(1/2)sqrt(15*(5/8)3)+....+(1/2)sqrt(15*(5/8)n-1)
=-(1/4)sqrt15+(1/2)sqrt15 +(1/2)sqrt(15)sqrt(5/8))+(1/2)sqrt(15)sqrt(5/8)2)+......+(1/2)sqrt(15)sqrt(5/8)n-1)
=-(1/4)sqrt15 +(1/2)sqrt(15)[1+sqrt(5/8)+sqrt(5/8)2+sqrt(5/8)3+........+sqrt(5/8)n-1]
sum of nterms of geometric progression is a(1-rn)/(r-1)
=-(1/4)sqrt15 +(1/2)sqrt(15)[1(1 -(sqrt(5/8))n)/(1-sqrt(5/8))]
=-(1/4)sqrt15 +(1/2)sqrt(15)[(1 -(sqrt(5/8))n)/(1-sqrt(5/8))]
=(sqrt15)[-(1/4) + (1/2)[(1 -(sqrt(5/8))n)/(1-sqrt(5/8))]]
time at nth bounce =
=(sqrt15)[-(1/4) + (1/2)[(1 -(sqrt(5/8))n)/(1-sqrt(5/8))]]
c) total time ==>n =infinity
=(sqrt15)[-(1/4) + (1/2)[(1 -(sqrt(5/8))infinity)/(1-sqrt(5/8))]]
=(sqrt15)[-(1/4) + (1/2)[(1 -0)/(1-sqrt(5/8))]]
=(sqrt15)[-(1/4) + (1/2)[1/(1-sqrt(5/8))]]
=2.1374(sqrt15)
=8.278seconds
total time =8.28 seconds approximately
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