Question

When a rubber ball is dropped on a particular surface , It bounces back to 8/10 of drop height. The ball's height after bounce x is * (8/10) ^ x times its drop height . What fraction of its drop height will the ball bounce up to after its fourth bounce ? Show your work

Answer 1

The fraction of its drop height after the fourth bounce is

`\displaystyle (16)/(625)`

Explanation:

Let's suppose the rubber ball is dropped from a height H0. Every time the ball bounces back it reaches 8/10 of the initial drop height.

The ball's height H(x) after bounce number x is:

`\displaystyle H(x)=H_o\left((8)/(20)\right)^x`

Simplifying:

`\displaystyle H(x)=H_o\left((2)/(5)\right)^x`

After the fourth bounce, the height is:

`\displaystyle H(4)=H_o\left((2)/(5)\right)^4`

`\displaystyle H(4)=H_o(2^4)/(5^4)`

`\displaystyle H(4)=H_o(16)/(625)`

The fraction of its drop height after the fourth bounce is

`\mathbf{\displaystyle (16)/(625)}`