Question

A superball that rebounds 3/10 of the height from which it fell on each bounce is dropped from 38 meters. ?

How high does it rebound, in meters, on the 8 th bounce?
How far does it travel, in meters, before coming to rest?

Answer 1

`8` th bounce, height = `38(3/10)^8` distance traveled to rest = `70.5714`

Explanation:

The sequence of heights after leaving is
`38{1,3/10,(3/10)^2,(3/10)^3,...,(3/10)^8,...,(3/10)^n}`
The space `d` traveled is given by
`d=2times 38{1+3/10+(3/10)^2+(3/10)^3+...+(3/10)^n}-38`
Now using the polynomial identity
`(1-x^{n+1})/(1-x)=1+x+x^2+x^3+...+x^n`
`1+3/10+(3/10)^2+(3/10)^3+...+(3/10)^n= (1-(3/10)^{n+1})/(1-(3/10))`
Supposing that `n->infty`,
then `(3/10)^{n+1}->0` because `(3/10)<1`
So we get
`1+3/10+(3/10)^2+(3/10)^3+...+(3/10)^n+...= 1/(1-(3/10)) = 10/7`
Finally putting all together
`d = 2 times 38 times 10/7 -38 = 70.5714`