Explain the conservation of mechanical energy using the example of a falling ball.

(N/A) Consider a ball of mass $m$ falling from the top of a cliff of height $H$,as shown in the figure.
For heights $H$,$h$,and $0$ (at the ground),the total mechanical energies $E_H$,$E_h$,and $E_0$ are calculated as follows:
At height $H$,the total mechanical energy is:
$E_H = mgH + \frac{1}{2}mv^2$
Since the velocity at the maximum height is $v = 0$,
$\therefore E_H = mgH$
At height $h$,the total energy is:
$E_h = \text{Potential Energy} + \text{Kinetic Energy}$
$E_h = mgh + \frac{1}{2}mv_h^2$ (where $v_h$ is the velocity of the ball at height $h$)
At the ground (height $0$),the total energy is:
$E_0 = \frac{1}{2}mv_f^2$
where $v_f$ is the final velocity of the ball when it hits the ground.
By the law of conservation of energy,$E_H = E_0$:
$mgH = \frac{1}{2}mv_f^2$
$\therefore v_f = \sqrt{2gH}$
Also,from the conservation of mechanical energy between height $H$ and height $h$:
$E_H = E_h$
$mgH = mgh + \frac{1}{2}mv_h^2$