Question

Show that the total mechanical energy of a body falling freely under gravity is conserved.

Answer 1

For freely falling body, V^2-U^2= 2gh, where V is final speed and U is initial speed. 
At the top all energy is potential energy in form of m*g*h. 
So U=0 and at any instant V^2=2gh. Where h is distance from top 
hence at any height h=(V^2)/2g. 
Thus mgh= mg(V^2)/2g = (1/2)(mV^2) which is the kinetic energy