Answer
$70$ meters
Work Step by Step
Vertical distance travelled in $I$ strike $=10$ m
Vertical distance travelled in $II$ strike $=2\times \frac{3}{4}\times 10 = \frac{15}{2}$ m
Vertical distance travelled in $III$ strike $=2\times \frac{3}{4}\times \frac{3}{4}\times 10 = \frac{45}{8}$ m
Vertical distance travelled in $IV$ strike $=2\times \frac{3}{4}\times \left(\frac{3}{4}\right)^2\times 10 = \frac{135}{32}$ m
and so on...
Assuming that ball bounces infinitely, the total distance travelled (in meters) will be
\begin{align*}
S&=10+2\times \frac{3}{4}\times 10+2\times \frac{3}{4}\times \left(\frac{3}{4}\right)^2\times 10+2\times \frac{3}{4}\times \left(\frac{3}{4}\right)^3\times 10+\cdots\\
&=10+20\left(\frac{3}{4}\right)+20\left(\frac{3}{4}\right)^2+20\left(\frac{3}{4}\right)^3+\cdots\\
&=10+20\left(\frac{3}{4}+\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+\cdots\right)\\
&=10+ \sum_{k=1}^\infty 20\left(\frac{3}{4}\right)^k\
\end{align*}
The series $\sum_{k=1}^\infty 20\left(\frac{3}{4}\right)^k$ is a geometric series $\sum_{k=1}^\infty ar^k$ with $a=20\times\frac{3}{4}$ and $r=\frac{3}{4}$. Its value
\begin{align*}
&=\frac{a}{1-r}\\
&=\frac{20\times\frac{3}{4}}{1-\frac{3}{4}}\\
&=60.
\end{align*}
So the total vertical distance travelled is $10+60=70$ meters.
The total vertical distance travelled is $\textbf{70 meters.}$
