91Ó°ÊÓ

The initial height is 20 feet.

When the ball strikes the ground then it bounces up to three-quarters of the previous height.

When the ball strikes the ground for the first time, then it bounces up to $20×\frac{3}{4}$feet.

When the ball strikes the ground for the first time, then it bounces up to $20×{\left(\frac{3}{4}\right)}^2$feet.

When the ball strikes the ground for the first time, then it bounces up to $$\begin{gathered} 20×{\left(\frac{3}{4}\right)}^3=20×\frac{27}{64} \\ ≈8.44 \end{gathered}$$

The sequence formed is:

$20\left(\frac{3}{4}\right),20{\left(\frac{3}{4}\right)}^2,20{\left(\frac{3}{4}\right)}^3................$

This is a geometric sequence whose

first term is $a=20\left(\frac{3}{4}\right)$

Common ratio is:$\frac{3}{4}$

So when the ball strikes $n$times then,

The height is:

$20\left(\frac{3}{4}\right){\left(\frac{3}{4}\right)}^{n-1}=20{\left(\frac{3}{4}\right)}^n$

Let ball bounces $t$times.

Then $$\begin{gathered} 20\left(\frac{3}{4}\right){\left(\frac{3}{4}\right)}^{t-1}<6 \\ 20{\left(\frac{3}{4}\right)}^t<6 \\ {\left(\frac{3}{4}\right)}^t<\frac{6}{20} \\ {(0.75)}^t<0.3 \end{gathered}$$

Now take log on both sides:

$$\begin{gathered} {\log }_{10}{(0.75)}^t={\log }_{10}(0.3) \\ t{\log }_{10}(0.75)={\log }_{10}(0.3) \\ t=\frac{{\log }_{10}(0.3)}{{\log }_{10}(0.75)} \\ t=4.18 \end{gathered}$$

So for four times ball needs to strike the ground before it bounce is less than 6 inches.

When the ball stops bouncing in infinity times.

So total distance traveled by the ball, before it stops bouncing is:

So, $$\begin{gathered} D=\frac{20\left({\displaystyle \frac{3}{4}}\right)}{1-{\displaystyle \frac{3}{4}}} \\ =\frac{15}{{\displaystyle \frac{1}{4}}} \\ =15×4 \\ =60 \end{gathered}$$