Question 552039
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If a ball rebounds three-fifths as far as it falls, how far will it (vertically) travel before coming to rest 
if dropped 14 feet?
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        The solution in the post by @Theo is incorrect.

        I came to bring a correct solution.



<pre>
First rebound is  {{{a[1]}}} = {{{(3/5)*14}}} = {{{42/5}}} ft.


The heights of rebounds that follow make geometric progression with the common ratio of  r = 3/5.


The infinite sum of all the rebound heights series is


    {{{a[1]/(1-r)}}} = {{{((42/5))/(1-3/5)}}} = {{{((42/5))/((2/5))}}} = {{{42/2}}} = 21.


This value, 21 ft, we should double, counting each fall-rebund, and add the initial height of 14 ft.


So, the final answer for the total travel is 14 + 2*21 = 14 + 42 = 56 ft.
</pre>

Solved correctly.