Explain qualitatively how the total energy of a freely falling body is conserved.

Let us consider an object possesing mass 'm' falling freely from a height 'h'. Initially, 

P.E of the body = mgh 

K.E of the body = 0 [ body is at rest] 

Therefore, 

Total energy of the body = mgh+0 = mgh

As the process tails, its potential energy will change into kinetic energy.

At a given instant, v is the velocity of the moving body. 

So, K.E of the body = $\frac{1}{2}m{v}^2$ 

During the fall, the potential energy would decrease while the kinetic energy would increase. When the object is about to reach the ground, h = 0 and velocity will be maximum. Therefore, the kinetic energy would be maximum and potential energy minimum. However, the total energy of the system, that is, the sum of potential and kinetic energy remains constant. 

That is, 

                $\mathrm{mgh} + \frac{1}{2}{\mathrm{mv}}^2 = \mathrm{constant}$

Thus, the law of conservation of energy is satisfied.