Question

A ball is dropped straight down from a height of `12` feet. Upon hitting the ground it bounces back `1/3` of the distance it fell. How far will the ball travel (both upward and downward) before it comes to rest?

Answer 1

The ball will travel 24 feet.

Explanation:

This problem requires the consideration of infinite series. Consider the actual behavior of the ball:

First the ball falls 12 feet.
Next the ball bounces up `12/3 = 4` feet.
The ball then falls the 4 feet.

On each successive bounce, the ball travels
`2*12/(3^n) = 24/3^n` feet, where `n` is the number of bounces

Thus, if we imagine that the ball starts from `n = 0`, then our answer can be gained from the geometric series:
`[sum_(n=0)^infty 24/3^n] - 12`

Note the `-12` correction term, this is because if we start from `n=0` we are counting a 0th bounce of 12 feet up and 12 feet down. In reality the ball only travels half of that, as it starts in midair.

We can simplify our sum to:
`[24sum_(n=0)^infty 1/3^n] - 12`

This is just a simple geometric series, which follows the rule that:
`lim_(n->infty)sum_(i=0)^n r^i = 1/(1 - r)`
As long as `|r| < 1`

This yields a simple solution to our problem:
`[24sum_(n=0)^infty 1/3^n] - 12 = 24*1/(1-1/3) - 12`
`= 24*1/(2/3) - 12 = 24*3/2 -12`
`= 36 - 12 = 24` feet.