An object is dropped from height H on ground. Every time it hits ground looses 50 percent of KE find total distance covered t->infinity

When the object falls through a height of h , its velocity is given by v^2 = 2 x g x h 

Hence its KE is 1/2 x m x 2gh = mgh

Given that it loses half of its KE on impact. So it rises with an initial KE of 1/2 mgh

So 1/2 mgh = 1/2 m u^2 where u is the velocity with which it rises. Or u^2 = gh

Now it will rise to a height of say s

So 0 = gh- 2 g x s and that gives s = h/2

This means that the object will rise to half of the height from which it fell after the first fall.

Similarly it will rise to a height of h/4 after the second fall and so on.

so the total distance covered will be h( 1 + 1/2 + 1/4 + 1/8+ ....)

In other words the total distance will be h (2) = 2h              Correct option is A