A particle falls from a height 'h' upon a fixed horizontal plane and rebounds. If 'e' is the coefficient of restitution the total distance travelled before rebounding has stopped is :-
A. $h\left( {\dfrac{{1 + {e^2}}}{{1 - {e^2}}}} \right)$
B. $h\left( {\dfrac{{1 - {e^2}}}{{1 + {e^2}}}} \right)$
C. $\dfrac{h}{2}\left( {\dfrac{{1 - {e^2}}}{{1 + {e^2}}}} \right)$
D. $\dfrac{h}{2}\left( {\dfrac{{1 + {e^2}}}{{1 - {e^2}}}} \right)$

Hint:In order to solve this question, we will find the distance covered by the particle each time after it hits the horizontal plane and gets rebound and then we will add the all distance covered by the body and we will count up-to infinity.

Formula used:
Newton’s equations of motion are ${v^2} - {u^2} = 2gS$ and for the vertical falling body and if it’s free fall then $u = 0$ and S is distance covered by the body in time t whose final velocity is v.

Complete step by step solution:
According to the question, when particle hits first time the floor then its velocity will be
${v^2} = 2gh \\
\Rightarrow v = \sqrt {2gh} $
where h is distance covered by it which is $h = \dfrac{{{v^2}}}{{2g}}$. and after bouncing from the plane its velocity will be $u = ev$ and it will cover a distance of ${h_1} = \dfrac{{{e^2}{v^2}}}{{2g}}$ while going upwards so before coming to again downwards to hit the plane it will cover a total distance of $2{h_1} = 2\dfrac{{{e^2}{v^2}}}{{2g}}$.

So on upcoming bounces of the particle the distance will form a series of the form,
$\dfrac{{{v^2}}}{{2g}},2\dfrac{{{e^2}{v^2}}}{{2g}},2\dfrac{{{e^4}{v^2}}}{{2g}},.....\infty $
So, total distance will be,
$S = \dfrac{{{v^2}}}{{2g}} + 2\dfrac{{{e^2}{v^2}}}{{2g}} + 2\dfrac{{{e^4}{v^2}}}{{2g}} + ..... + \infty \\
\Rightarrow S = \dfrac{{{v^2}}}{{2g}} + 2\dfrac{{{v^2}}}{{2g}}[{e^2} + {e^4} + .... + \infty ] \\ $
And here, $[{e^2} + {e^4} + .... + \infty ]$ is a infinite geometric series of first term $a = {e^2}$ and common ratio ${e^2}$ and sum of geometric series is given by,
${S_\infty } = \dfrac{a}{{1 - r}}$

On putting the values we get,
$[{e^2} + {e^4} + .... + \infty ] = \dfrac{{{e^2}}}{{1 - {e^2}}}$
So, we have
\[S = \dfrac{{{v^2}}}{{2g}} + 2\dfrac{{{v^2}}}{{2g}}[{e^2} + {e^4} + .... + \infty ] \\
\Rightarrow S = \dfrac{{{v^2}}}{{2g}} + 2\dfrac{{{v^2}}}{{2g}}[\dfrac{{{e^2}}}{{1 - {e^2}}}] \\
\Rightarrow S = \left( {\dfrac{{1 + {e^2}}}{{1 - {e^2}}}} \right)\dfrac{{{v^2}}}{{2g}} \\
\therefore S = h\left( {\dfrac{{1 + {e^2}}}{{1 - {e^2}}}} \right) \\ \]
Hence, the correct answer is option A.

Note: It should be noted that, we have taken infinite series because we know that due to gravitational force after very long period of time the particle will sure comes to at rest but we are not sure after how many bounces that’s why we have taken infinite number of distances.