An object is dropped from a height h from the ground . Every time it it hits ground it loses 50 % of its kinetic energy . The total distance covered as ` t to oo` is :

Options

A. 3h
B. `oo`
C. `5/3` h
D. `8/3` h

Answer

A

Explanation

Let mass of object is m , and droped from height h on the ground , energy is converted from P.E to K.E Kinetic energy of object = potential energy of object .
K.E = mg(h)
This object hite the ground but loose 50 % of initial K.E
K.E =mg(h/2)
Every time object half of the height from which it was dropped . Total distance covered by object hiting the ground toll `t to oo`
`H = h+2 (h/2 +h/(2^(2)) +....+h/(2^(n))) = h+h (1+1/2 +1/(2^(2))+....oo)`
`= h + h (1/(1-1/2))`
H = 3 h