Shyam drops a ball of 100 g from a building of height 10 m. What will be its kinetic energy at the height of 4 m? What will happen to its total mechanical energy? Give reasons to justify your answer.

Potential energy, PE = mgh = 0.1 X10 X10 = 10J

When body reaches at height 6 m , in velocity is given by

{{v}^{2}} = {{u}^{2}} + 2gh

{{v}^{2}} = 0 + 2 x 10 x 6

Now, its kinetic energy at this point is given by,

= 1/2 m${{v}^{2}}$ = 1/2 x 0.1 x 4 x 30 = 6 J

As, mechanical energy = KE + PE [at highest point]

Mechanical energy = 0+10 J = 10J

Mechanical energy at 4 m height

= KE+ PE= 6 + 0.l x 10x 4= 10J

Hence, it is seen that total mechanical energy always remains constant during the motion.