We know that each time the ball bounces it travels a distance of 3/4 the original height, that means that for each bounce the ball will rise:
![a_n=(3)/(4)a_(n-1)](https://img.qammunity.org/2023/formulas/mathematics/college/w1qfyrklqzs59e39gbrhkvdk1eytsany3n.png)
where a_n is the height of the nth bounce and a_(n-1) is the bounce n-1 (the previous one).
Now, from this formula we know that:
![a_1=(3)/(4)a_0=(3)/(4)\cdot6=(9)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/sdkox62nav4n66hy6alxv2tlofu3z59yis.png)
Therefore the explicit formula for the height of the nth bounce is:
![\begin{gathered} a_n=(9)/(2)((3)/(4))^(n-1) \\ =(9)/(2)\cdot(4)/(3)((3)/(4))^n \\ =6((3)/(4))^n \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/w5zt20bfomczapzpz5gwo4ziughpsza19f.png)
Then:
![a_n=6((3)/(4))^n](https://img.qammunity.org/2023/formulas/mathematics/college/2kni7zu5p3nqztop9rmecs81oidhnaq0e7.png)
Now, to find the total distance it travels by the fifth bounce we follow:
The first time the ball hits the ground it has traveled a distance of 6 ft.
Now, for the second bounce we have:
![6((3)/(4))+6((3)/(4))=12((3)/(4))](https://img.qammunity.org/2023/formulas/mathematics/college/jbza4un5nv4mqb61h7yg14hj1vfxroa69k.png)
(one distance up and one distance down).
For the third bounce we have:
![6((3)/(4))((3)/(4))+6((3)/(4))((3)/(4))=12((3)/(4))^2](https://img.qammunity.org/2023/formulas/mathematics/college/ilge3zojf7ax7u19nbr2a426l8r5vt4f8l.png)
If we continue this process we notice that the ball will travel a total distance of:
![6+12((3)/(4))+12((3)/(4))^2+12((3)/(4))^4+\ldots\text{..}](https://img.qammunity.org/2023/formulas/mathematics/college/kbcviudki196nh2ko2xp4zbh8ia8a60lmy.png)
but this can be written (for an infinity number of bounces) as:
![\begin{gathered} 6+12\sum ^(\infty)_(n\mathop=0)((3)/(4))^(n+1) \\ =6+12((3)/(4))\sum ^(\infty)_(n\mathop=0)((3)/(4))^n \\ =6+9\sum ^(\infty)_(n\mathop=0)((3)/(4))^n \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/a88jeupa8mr3d2h7n88n23o2n7s6jcfl7p.png)
As we said this is for an infinity of bounces if we like only five bounces then we have:
![6+9\sum ^5_(n\mathop=0)((3)/(4))^n](https://img.qammunity.org/2023/formulas/mathematics/college/6yb50zvqj6oyhe0pm5bctczahyn7afpb56.png)
Now to find the distance in this case we do the sum and we get appoximately 35.5928 feet.