For the situation depicted in given figure, calculate the mechanical energy of the ball just before it hits the ground and show that even at this position, it is mgh.[Pg. No. 129]

At point A, the ball is at height $h$. So, its mechanical energy is:

$$\begin{aligned} \text{Mechanical Energy} &= \text{Potential Energy} + \\ &\quad \text{Kinetic Energy} \\ &= mgh + 0 = mgh \end{aligned}$$

Just before the ball hits the ground, its height becomes zero. Hence, potential energy becomes zero, but the ball gains kinetic energy.

So, Mechanical Energy = 0 + Kinetic Energy

The kinetic energy gained by the ball is equal to the loss in potential energy, i.e.,

$$\text{Kinetic Energy} = mgh$$

Therefore, Mechanical Energy = $mgh$

Hence, even just before hitting the ground, the mechanical energy of the ball remains $mgh$.