A Ball That Bounces Forever (in Finite Distance)

A dropped ball rebounds to a fixed fraction of its height every bounce, so it bounces infinitely many times. Yet the total distance it travels is finite: the bounce heights form a geometric series, and that sum has a clean formula.

How it works

Series and geometric series

In one sentence

A series $\sum a_n$ converges when its partial sums $s_N = \sum_{n=1}^N a_n$ approach a limit; the geometric series $\sum_{n=0}^\infty ar^n$ converges to $\frac{a}{1-r}$ when $|r| < 1$ and diverges otherwise.

General formula

{\htmlData{sym=0}{D}} = {\htmlData{sym=1}{h}} + 2{\htmlData{sym=1}{h}}{\htmlData{sym=2}{r}} + 2{\htmlData{sym=1}{h}}{\htmlData{sym=2}{r}}^2 + \cdots = {\htmlData{sym=1}{h}} + \frac{2{\htmlData{sym=1}{h}}{\htmlData{sym=2}{r}}}{1-{\htmlData{sym=2}{r}}}, \qquad 0 < {\htmlData{sym=2}{r}} < 1

Each bounce adds a fixed fraction of the previous one, so what's left to add shrinks by that fraction every step and the total closes in on a limit.

What the symbols mean

$D$
Total distance
Everything the ball travels, up and down, over all bounces.
$h$
Drop height
Where the ball starts.
$r$
Rebound ratio
Each bounce reaches this fraction of the last one's height.

The ball falls $h$ , then rises and falls $hr$ (twice), then $hr^2$ (twice), and so on.
The rebounds form a geometric series, $2hr(1 + r + r^2 + \cdots)$ .
For $|r| < 1$ , $1 + r + r^2 + \cdots = \frac{1}{1-r}$ : the partial sums approach a limit.
So the total distance is $h + \frac{2hr}{1-r}$ . The bounce times shrink geometrically too, so the ball even stops in finite time.

Learn the math

Series: The BasicsCalc II · Lesson 26, Sequences and Series Geometric and Telescoping SeriesCalc II · Lesson 28, Sequences and Series