Show that the sum of kinetic energy and potential energy (ie, total mechanical energy) is always conserved in the case of a freely falling body under gravity (with air resistance neglected from a height by finding it when (i) the body is at the top c) the body has fallen a distance in the body has reached the ground.

Explanation

Consider a body of mass m be dropped from A
Case -1( at A) PE= mgh
K.E= 0
Total energy = mgh
Case -2 ( At B)
applying `v^(2)= u^(2)+2as`
velocity at B can be obtained as
`v^(2)= 0^(2)+2gx`
`K.E= 1//2mv^(2)= 1//2 m.2gx`
`K.E= mgx`
`P.E= mg(h-x) `
`T.E= mgx+mg(h-x)= mg(x+h-x)` ....(ii)
T.E= mgh

Velocity can be obtained as `v^(2)= (0)^(2)+2gh`
`K.E= (1)/(2)mv^(2)= (1)/(2)m(2gh)`
`K.E= mgh`
P.E= 0
T.E= mgh .....(iii)
From (i), (ii) and (iii) we can conclude that total energy of a freely falling body remains same.