Answer

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**Hint**We are here asked to find the relationship between $h$ and $d$. Also we are given with the coefficient of restitution. Thus, it would be easier to go through the path of restitution analysis.

Formulae Used:

${v^2} - {u^2} = 2ah$

Where, $v$ is the final velocity of the particle, $u$ is the initial velocity of the particle, $a$ is the acceleration on it and $h$ is the height of the particle.

$e = \dfrac{{{v_2}}}{{{v_1}}}$

Where, $e$ is the coefficient of restitution, ${v_2}$ is the velocity of the particle after collision and ${v_1}$ is the velocity of the particle before collision.

**Complete Step By Step Solution**

Here,

For the first bounce of the ball on the top most step,

${v_1}^2 - {0^2} = 2ah$

We took ${u_1} = 0$ as the initial velocity of the ball was zero.

Thus, we get

$2ah = {v_1}^2 \cdot \cdot \cdot \cdot (1)$

Now,

After the first bounce, the new height of the ball is $(h - d)$,

Thus, we get

$2a(h - d) = {v_2}^2 \cdot \cdot \cdot \cdot (2)$

Now,

Applying$\dfrac{{Equation(1)}}{{Equation(2)}}$, we get

$\dfrac{h}{{h - d}} = \dfrac{{{v_1}^2}}{{{v_2}^2}}$

After further evaluation, we get

$\dfrac{{{v_2}}}{{{v_1}}} = \sqrt {\dfrac{{h - d}}{h}} $

But,

$e = \dfrac{{{v_2}}}{{{v_1}}}$

Thus, equating both, we get

$e = \sqrt {\dfrac{{h - d}}{h}} $

Further, we get

${e^2} = \dfrac{{h - d}}{h} \Rightarrow h{e^2} = h - d \Rightarrow h = \dfrac{d}{{1 - {e^2}}}$

**Hence, the correct option is (A).**

**Note**We have used the route of using the fundamental formula for coefficient of restitution as that allows us to directly relate all the given parameters. Moreover, in the given case it is given that the collision is head on which means completely elastic, but if the collision was inelastic, then the evaluation will be somewhat different.

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