tion:HELLO friends PLS find the answer below law of conservation of energy states that the energy can neither be created nor be DESTROYED but can be transformed from one form to another....let us now PROVE that the above law holds good in the case of a freely falling body....let a body of mass m placed at a HEIGHT h above the ground, start falling down from rest...in this case we have to show that the total energy ( potential energy + kinetic energy) of the body at A, B and C remains constant i.e, potential energy is completely transformed into kinetic energy...AT A-- potential energy=mgh kinetic energy= 1/2mv square = 1/2*m* 0 kinetic energy= 0 ( the velocity is zero as the object is initially at rest)./// total energy at a = potential energy + kinetic energy = mgh + 0 Total energy at a= mgh ............1eq.....AT B -- potential energy= mgh=mgh( h- x ) height from the ground is ( h-x) potential energy= mgx kinetic energy= 1/2mv square the body covers the distance x with a velocity vv. we make use of the third equation of motion to obtain velocity of the body.....v2- u2u2= 2a here, u=0, a=g, s=xv2- 0= 2 ax here, u= 0 ,a= g and s= xv2 - 0= 2gx..v2 = 2gx kinetic energy= 1/2mv square= 1/2m2gx..kinetic energy= mgxTotal energy at b= potential energy+ kinetic energy= mgh- mgx+ mgx Total energy at B = mgh.....2eq....AT C :- potential energy = m x g x 0 ( h = 0 ) potential energy = 0 kinetic energy =1/2mv2 the distance covered by the body is hv2 - u2 = 2as here, u = 0 ,a= g, and s= h v2 - 0= 2ghv2 = 2gh kinetic energy = 1/2mv2 = 1/2* 2gh total energy at C = potential energy + kinetic energy = 0 + mgh Total energy at C = mgh....3eq....its clear from equation 1,2 and 3 that the total energy of the body remains constant at every point..thus, we conclude that law of conservation of energy holds good in the case of a freely falling body....hope this helps uh......plzz mark it as brainliest....click on thnxxxx rate this answer......follow me guys..........