Geometric Series-Bounce Height of a Ball

Question

A ball is dropped from a height of $20 \, \mathrm{m}$. It rebounds to a height of $16 \, \mathrm{m}$ and continues to rebound to eight-tenths of its previous height for subsequent bounces.

Calculate the total distance the ball travels before it comes to rest.

How do you answer this question?

I'm using the formula $$ S_n = \frac{a(1-r^n)}{1-r}, \qquad r\lt 1 $$ I believe that the variables: $$ \begin{align} a &= 20 \\ r &= 0.8 \\ n &= \text{undefined} \end{align} $$ The answer is meant to equal $100 \, \mathrm{m}$ so, $S_n = 100 \, \mathrm{m}$.

Please help.

Thank you in advance.

Answer 1

You need the sum of the infinite geometric series given by $$S=20+\frac{8}{10}\times20+\left(\frac{8}{10}\right)^2\times20+\cdots$$

For a geometric series $$S=a+ar+ar^2+\cdots$$ The $N^{th}$ partial sum is given by $$S_N=a+ar+ar^2+\cdots+ar^{N-1}=\frac{a(1-r^N)}{1-r}$$

Then, $$S=\lim_{N\to\infty}S_N=\lim_{N\to\infty}\frac{a(1-r^N)}{1-r}$$

If $|r|<1$, the limit exists and is given by, $$S=\frac{a}{1-r}$$ Hence, here $$S=\frac{20}{1-\frac{8}{10}}=100$$

Answer 2

Let's say that at iteration $n$ the ball goes down $d_n$ meters, and then goes up $u_n$ meters, for all $n \ge 0$. We are told that $d_0 = 20$ and that $u_n = \frac 8 {10} d_n = \frac 4 5 d_n$. It is also clear that $d_{n+1} = u_n$ for all $n \ge 0$. The total distance is

$$(d_0 + u_0) + (d_1 + u_1) + \dots = \sum _{n \ge 0} \left( d_n + \frac 4 5 d_n \right) = \frac 9 5 \sum _{n \ge 0} d_n \ .$$

Now, using the relations from the first paragraph, $d_{n+1} = u_n = \frac 4 5 d_n$, whence we get that $d_{n+1} = \left( \frac 4 5 \right) ^{n+1} d_0$ for all $n \ge 0$, whence it is immediate that $d_n = \left( \frac 4 5 \right) ^n d_0$ for all $n \ge 0$. Using this back in our sum, the total distance becomes

$$\frac 9 5 \sum _{n \ge 0} \left( \frac 4 5 \right) ^n d_0 = \frac 9 5 \frac 1 {1 - \frac 4 5} d_0 = 180 \ .$$