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!