The total amount ofmechanical energy, in a closed system in the absence of dissipative forces (e.g. friction, air resistance), remains constantAt Point A:At point A, the ball is stationary; therefore, its velocity is zero.Therefore, kinetic energy, T = 0 and potential energy, U = mghHence, total mechanical energy at point A isE = T + U = 0 + mgh = mgh... (i)At Point B :Suppose the ball covers a distance x when it moves from A to B. Let v be the velocity of the ball point B. Then by the equation of motion v^2-u^2 = 2aS, we havev^2 - 0 = 2gx or v^2 = 2gx Therefore,Kinetic energy, T = 1/2 mv^2 = 1/2 x m x (2gx)= mgxAnd Potential energy, U = mg (h - x)Hence, total energy at point B isE = T + U = mgx + mg(h-x) = mgh ...(ii)At Point C :Suppose the ball covers a distance h when it moves from A to C. Let V be the velocity of the ball at point C just before it touches the ground. Then by the equation of motion v^2 - u^2 = 2aS, we have V^2 - 0 = 2gh or V^2 = 2gh.Therefore,Kinetic energy,T = 1/2 mV^2 = 1/2 x m x (2gh) = mghand Potential energy, U = 0Hence, total energy at point E = T + U= mgh + 0 = mgh... (iii)Thus, it is clear from equations (i), (ii) and (iii), that the total mechanical energy of a freely falling ball remains constant.