Answer
$\text{6 m}$
Work Step by Step
This is the form of an infinite geometric series, in which the first term is ${{a}_{1}}=12$ and the common ratio is $r=\frac{1}{3}$.
The limit and sum of the infinite geometric series is given by
${{s}_{\infty }}=\frac{{{a}_{1}}}{1-r},\text{ }\left| r \right|<1$
Where ${{a}_{1}}=$first term and $r=$common ratio.
If $\left| r \right|<1$, then the limit exists, and if $\left| r \right|\ge 1$, no limit exists.
Since, $\left| r \right|=\frac{1}{3}<1$, the limit exists.
Therefore, the sum of an infinite geometric series is
${{s}_{\infty }}=\frac{{{a}_{1}}}{1-r},\text{ }\left| r \right|<1$ …… (1)
Substitute the value of ${{a}_{1}}=12$ and $r=\frac{1}{3}$ in equation (1)
$\begin{align}
& {{s}_{\infty }}=\frac{{{a}_{1}}}{1-r} \\
& =\frac{12}{1-\frac{1}{3}} \\
& =\frac{12\times 3}{3-1} \\
& =18
\end{align}$
Therefore,
${{s}_{\infty }}=18.$
The total distance is $18\text{ m}$, and the fall distance is $12\text{ m}$; therefore, the rebound distance is
$18-12=6\text{ m}$.
Thus, the total rebound distance is $\text{6 m}$.
