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

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!

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