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

Answer:

The fraction of its drop height after the fourth bounce is

$\displaystyle \frac{16}{625}$

Step-by-step 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:

$\displaystyle H(x)=H_o\left(\frac{8}{20}\right)^x$

Simplifying:

$\displaystyle H(x)=H_o\left(\frac{2}{5}\right)^x$

After the fourth bounce, the height is:

$\displaystyle H(4)=H_o\left(\frac{2}{5}\right)^4$

$\displaystyle H(4)=H_o\frac{2^4}{5^4}$

$\displaystyle H(4)=H_o\frac{16}{625}$

The fraction of its drop height after the fourth bounce is

$\mathbf{\displaystyle \frac{16}{625}}$