Answer:
The
mechanical energy (kinetic energy + potential energy) of a freely falling object
remains constant. It may be shown by calculation as follows:
Suppose a body of mass m falls from point A, which is at height 'H' from
the surface of earth. Initially at point A, kinetic energy is zero and the body
has only potential energy.
Total energy of body at point A = Kinetic energy +
Potential energy = 0 + mgH = mgH
Suppose during fall, the body is at position B. The body
has fallen at a distance x from its initial position. If velocity of body at B
is v, then from formula 
we have
Kinetic energy of body at
point 
Potential energy of body at point B = mg (H -
Total energy of body at
point B = Kinetic energy + Potential energy
= mgx + mg (H - x) = mgH ...(ii)
Now suppose the body is at point C, just above the surface
of earth (i.e., just about to strike the earth). Its potential energy is zero.
The height by which the body falls = H
If v is velocity of body at C, then from formula
v 2 = u 2 + 2as
We have 
So, 
Kinetic energy of body at
position 
Total energy of body at
C
= Kinetic energy + Potential energy
...(iii)
Thus, we see that the sum of kinetic energy and potential
energy of freely falling body at each point
remains constant.
Thus, under force of gravity, the total mechanical energy
of body remains constant.
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