Question 1099692
<pre>
The ball goes down and then up. So think of it an an infinite
series of "down-ups".  The first "down-up" consists of the first 
"down" of 9 meters plus the following "up" which is one-third of 
9 meters or 3 meters, which is 9+3 = 12 meters for the first "down-up".
Then each "down-up" after that is one-third of the previous "down-up".

So we use:

{{{S[infinity]}}}{{{""=""}}}{{{a[1]/(1-r^"")}}}

Where a<sub>1</sub> = 12 and {{{r=1/3}}}

{{{S[infinity]}}}{{{""=""}}}{{{12^""/(1-1/3)}}}

{{{S[infinity]}}}{{{""=""}}}{{{12/(2/3)}}}{{{""=""}}}{{{12*(3/2)}}}{{{""=""}}}{{{18}}}

The ball travels 18 meters.

Edwin</pre>