Infinite geometric series word problem

Sep 2015
11
0
Ohio
Q: A ball is dropped from a height of 32 ft. Each time it strikes the ground it rebounds 3/8ths of the distance from which it had fallen. Theoretically, how far will the ball travel before coming to a rest?

Because the ball rebounds up then falls back down again before rebounding a next time, I know I can't use the formula for sum of an infinte geometric series, a/(1-r). I'm not sure how to account for the rebounding in this problem.

I realized that the the solution to the question (70.4 ft) is equal to (11/8)(a/(1-r)), but I'm not sure where the 11/8 came from.

Any help is appreciated.
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
down travel ...

$\displaystyle 32\left[1 + \left(\frac{3}{8}\right) + \left(\frac{3}{8}\right)^2 + \left(\frac{3}{8}\right)^3 + \, ... \, \right]$

up travel ...

$\displaystyle 32\left[\left(\frac{3}{8}\right) + \left(\frac{3}{8}\right)^2 + \left(\frac{3}{8}\right)^3 + \, ... \, \right]$

up + down ...

$\displaystyle 32\bigg[1 + 2 \sum_{n=1}^\infty \left(\frac{3}{8}\right)^n \bigg]$
 
  • Like
Reactions: 1 person
Sep 2015
11
0
Ohio
That makes sense. So I only needed to add the sum of two infinite geometric sequences,

[(32)/(1-(8/3))] + [(12)/(1-(8/3))] = 70.4

Thanks.

(Sorry for bad formatting of equation, I'm using my phone's web browser)