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**Hint:**This question is based on the concepts of motion. We need to understand the equations of motion and their applications for a freely falling body. The coefficient of restitution is the ratio of final velocity to initial velocity for two objects after they collide. The restitution coefficient, denoted by the letter ‘e,' is a unitless quantity with values ranging from 0 to 1.

**Complete step by step answer:**

Let the velocity with which the ball strikes the plate be ${v_1}$. Let ${v_2}$ be the velocity of rebound. Let the height from which the ball is dropped be $H$.

From the equations of motion, we know that,

${v_1} = \sqrt {2gH} $

Here $g$ is acceleration due to gravity.

${v_2} = \sqrt {2g{h_1}} $

$

e = \dfrac{{{v_2}}}{{{v_1}}} = \sqrt {\dfrac{{{h_1}}}{H}} \\

{e^2} = \dfrac{{{h_1}}}{H} \\

{h_1} = {e^2}H \\

$

Similarly, we get,

${h_2} = {e^2}{h_1} = {e^4}H$and so on.

Therefore, total distance before it comes to stop,

$

y = H + 2{h_1} + 2{h_2} + 2{h_3} + ..... \\

= H + 2{e^2}H + 2{e^4}H + 2{e^6}H + ..... \\

= H[1 + 2{e^2}(1 + {e^2} + {e^4} + .....)] \\

= H[1 + \dfrac{{2{e^2}.1}}{{1 - {e^2}}}] \\

= H[\dfrac{{1 + {e^2}}}{{1 - {e^2}}}] \\

= 3[\dfrac{{1 + {{0.5}^2}}}{{1 - {{0.5}^2}}}] \\

= 5m \\

$

Thus, the ball will cover a distance of 5 meters before coming to rest.

**Note:**The coefficient of restitution gives us knowledge about the collision's elasticity. A perfectly elastic collision is described as one in which no overall kinetic energy is lost. The maximum coefficient of restitution for this form of collision is e = 1. A perfectly inelastic collision is one in which all of the kinetic energy is lost.

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