A steel ball is dropped on a hard surface from a height of `1m` and rebounds to a height of `64cm`. The maximum height attained by the ball after `n^(th)` bounce is (in m)

Options

A. (a) `(0.64)^(2n)`
B. (b) `(0.8)^(2n)`
C. (c) `(0.5)^(2n)`
D. (d) `(0.8)^n`

Answer

B

Explanation

`h_n=(e^(2n))hi`
Given that
`(64cm)=(e^2)(100cm)`
`:. e=0.8`
Substituting in Eq. (i) we have
`h_n=(0.8)^(2n)(1m)`
`=(0.8)^(2n)`