# Please give answer of c

If work done by external forces acting on a body is zero and all the internal forces are consider,then total mechanical energy of body remains conserved.

ui + Ki = uf +Kt

At point A,KEi = 0

ui=mgH

At point B,ut = 0

Ktf = 1/2 mv

^{2}

putting the value of V

K= 1/2m(2gh)

kEt = mgH ( TE = Kf + ut = mgH)

V2 = u2 + 2gH

V2 = 0 + 2gH

V = root over of 2gH

So,the initial potential energy was mgH and final kinetic energy also comes out to be mgH.

Hence,it is proved that the kinetic energy of freely falling body on reaching the ground is nothing but transformation of its initial potential energy.

Regards,

**
**