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# A ball is dropped from a height of h from ground. If the coefficient of restitution is e, the height to which the ball goes up after it rebounds for the nth time is

$(a)\;e^{2n}h \quad (b)\;he^n \quad (c)\;he^{2n} \quad (d)\;h/e^{2n}$

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A)
For an object bouncing off a stationary object, such as a floor, coefficient of restitution, $e = \sqrt{\large \frac{h'}{h}}$
$\Rightarrow$ The height to which it bounces $h' = e^2h$
Similarly, after it rebounds for the nth time, the height to which it rises $h' = e^{2n}h$
Answer: $e^{2n}h$

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A)

e=velocity of separation/velocity of approach

so,  in this case as ground is in rest so,

e=v/u only

every time it collides final velocity becomes initial thus power of 'e' increases.

for n collisions: final velocity(v)=(e^n)(initial velocity)

initial velocity=(2gh)^1/2

so, (v)=(e^n)((2gh)^1/2

thus, as it is uniform motion so,
2gH=v^2
2gH=(e^2n)2gh
thus,   H=(e^2n)h
so, option (a) is correct!