A rubber ball is bounced from a height of 4 metres and bounces continuously. Each successive bounce reaches a height that is half the previous height. If the pattern of the maximum height reached during each bounce continues,
After how many bounces will the ball reach a maximum height of 1/512 metres?
Previous part in
1 Answer | Add Yours
It is given that rubber ball bounce, half of the previous height. initially it bounces from 4m and starts to bounce.
Height of second bounce `= 4/2 =2m=4*1/2`
Height of third bounce `=2/2 =1 = 4*(1/2)^2`
Height of fourth bounce `=1/2 =4*(1/2)^3`
The motion of a bouncing of the follows a geometric series with a=4 and r=1/2
So the maximum height of the nth bounce would be;
`T_n = 4*(1/2)^(n-1)`
When the maximum height is 512 then T_n = 1/512
`1/512 = 4*(1/2)^(n-1)`
`1/2048 = (1/2)^(n-1)`
`1/2^11 = (1/2)^(n-1)`
`(1/2)^11 = (1/2)^(n-1)`
`11 = n-1`
`n = 12`
So the maximum height of 1/512 would be obtained when
n = 12 or in the 12th bounce.
We’ve answered 318,915 questions. We can answer yours, too.Ask a question