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:
[tex]a_n=\frac{3}{4}a_{n-1}[/tex]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:
[tex]a_1=\frac{3}{4}a_0=\frac{3}{4}\cdot6=\frac{9}{2}[/tex]Therefore the explicit formula for the height of the nth bounce is:
[tex]\begin{gathered} a_n=\frac{9}{2}(\frac{3}{4})^{n-1} \\ =\frac{9}{2}\cdot\frac{4}{3}(\frac{3}{4})^n \\ =6(\frac{3}{4})^n \end{gathered}[/tex]Then:
[tex]a_n=6(\frac{3}{4})^n[/tex]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:
[tex]6(\frac{3}{4})+6(\frac{3}{4})=12(\frac{3}{4})[/tex](one distance up and one distance down).
For the third bounce we have:
[tex]6(\frac{3}{4})(\frac{3}{4})+6(\frac{3}{4})(\frac{3}{4})=12(\frac{3}{4})^2[/tex]If we continue this process we notice that the ball will travel a total distance of:
[tex]6+12(\frac{3}{4})+12(\frac{3}{4})^2+12(\frac{3}{4})^4+\ldots\text{..}[/tex]but this can be written (for an infinity number of bounces) as:
[tex]\begin{gathered} 6+12\sum ^{\infty}_{n\mathop=0}(\frac{3}{4})^{n+1} \\ =6+12(\frac{3}{4})\sum ^{\infty}_{n\mathop=0}(\frac{3}{4})^n \\ =6+9\sum ^{\infty}_{n\mathop=0}(\frac{3}{4})^n \end{gathered}[/tex]As we said this is for an infinity of bounces if we like only five bounces then we have:
[tex]6+9\sum ^5_{n\mathop=0}(\frac{3}{4})^n[/tex]Now to find the distance in this case we do the sum and we get appoximately 35.5928 feet.
