$(i)$ State the law of conservation of energy.
$(ii)$ Show that the total mechanical energy of a freely falling body is conserved.

(N/A) $(i)$ The law of conservation of energy states that energy can neither be created nor destroyed,it can only be transformed from one form to another. For a system,the total energy remains constant if no external non-conservative forces act on it.
$(ii)$ Consider a body of mass $m$ falling freely under gravity from a height $h$ above the ground. Let the body be at point $A$ at height $h$,point $B$ at height $(h-x)$ after falling a distance $x$,and point $C$ just before hitting the ground.
At point $A$ (Height $h$): The body is at rest,so velocity $v = 0$.
Kinetic Energy $(T) = 1/2 mv^2 = 0$.
Potential Energy $(U) = mgh$.
Total Energy $(E) = T + U = 0 + mgh = mgh$ $....(i)$
At point $B$ (Height $h-x$): The body has fallen distance $x$. Using $v^2 - u^2 = 2as$,where $u=0, a=g, s=x$,we get $v^2 = 2gx$.
Kinetic Energy $(T) = 1/2 m(2gx) = mgx$.
Potential Energy $(U) = mg(h-x) = mgh - mgx$.
Total Energy $(E) = T + U = mgx + mgh - mgx = mgh$ $....(ii)$
At point $C$ (Height $0$): The body has fallen distance $h$. Using $v^2 - u^2 = 2as$,where $u=0, a=g, s=h$,we get $V^2 = 2gh$.
Kinetic Energy $(T) = 1/2 mV^2 = 1/2 m(2gh) = mgh$.
Potential Energy $(U) = mg(0) = 0$.
Total Energy $(E) = T + U = mgh + 0 = mgh$ $....(iii)$
From equations $(i), (ii),$ and $(iii)$,it is clear that the total mechanical energy of a freely falling body remains constant at all points.