Question

An object is dropped from a height h from the ground. every time it hits the ground it looses 50% of its kinetic energy. the total distance covered as

Answer 1

Let mass of body is m
Object is dropped from h height , so potential energy = kinetic energy
potential energy = mgh = kinetic energy ,K₁

Now, question said object 50% of kinetic in every hits the ground.
so, after hitting kinetic energy will remain = 1/2 of initial kinetic energy, K₂ = mgh/2
similarly 2nd hitting kinetic energy will remain = 1/2 of K₂ = mgh/4
......................................
nth hitting kinetic energy will remain = mgh/2ⁿ

We can see total distance covered by object = h +2( h/2 + h/4 + ......h/2ⁿ) [ here I multiplied with 2 because if object hit the ground and then rebound H height and then again falls to ground , means body take two times of H for 2nd hitting ]
This in GP progression ,
For infinite terms , Sn = a/(1 - r)
Here , a = h/2 and r = 1/2
So, total distance covered by object =h + 2[h/2/(1 - 1/2) = h + 2h = 3h