Question 978610
<pre>
Bounce Number         Rebound
      {{{1}}}          {{{(12/13)*8 = 96/13}}}
      {{{2}}}          {{{(12/13)*(96/13) = 1152/126}}}
      .
      .
      .
      {{{n}}}           {{{a*r^(n-1) = (96/13)(12/13)^(n-1)}}}

We set the nth bounce < 1

{{{(96/13)(12/13)^(n-1)}}}{{{""<""}}}{{{1}}}

Multiply both sides by {{{13/96}}}

{{{(12/13)^(n-1)}}}{{{""<""}}}{{{13/96}}}

Take logs of both sides:

{{{log(((12/13)^(n-1)))}}}{{{""<""}}}{{{log((13/96))}}}

{{{(n-1)log((12/13))}}}{{{""<""}}}{{{log((13/96))}}}

Divide both sides by {{{log((12/13))}}}

However since {{{log((12/13))=-.0347621063}}}

we are dividing an inequality through by a negative number
and that will reverse the inequality from < to >:


{{{n-1}}}{{{"">""}}}{{{log((13/96))/log((12/13))}}}

{{{n-1}}}{{{"">""}}}{{{24.97915041}}}

{{{n}}}{{{"">""}}}{{{25.97915041}}}

So the 26th rebound will be the first one less than 1 foot.

How far will the ball travel before it comes to rest on the ground? 

--------------------------------------

Theoretically it will never stop bouncing, but that's theoretically,
not actually.

However we treat it as an infinite series.

The first 8 feet that it fell must be added in separately.

It rebounded 96/13 feet and then fell 96/13 before it hit the ground
the second time.  So the first term is 2×(96/13) = 192/13 and the common
ratio is 12/13

{{{S[infinity]=a[1]/(1-r)=matrix(1,11,   (192/13),"÷",(1-12/13),""="",(192/13),"÷",(1/13),""="",(192/13)(13/1),""="",192)}}}

To that we must add the 8 feet it fell in the beginning:

Answer: 192+8 = 200 feet.

Edwin</pre>