Question

A ball of mass m is dropped vertically from a height h0 above the ground.

Answer 1

 

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