# 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?

May 16, 2016

$8$ th bounce, height = $38 {\left(\frac{3}{10}\right)}^{8}$ distance traveled to rest = $70.5714$

#### Explanation:

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