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?

1 Answer
JMicheli
Jul 12, 2016

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.

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