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...

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jeew-m | College Teacher | (Level 1) Educator Emeritus

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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.


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