A small disc A slides down with initial velocity equal to zero from the top of a smooth hill of height H having a horizontal portion. What must be the height of the horizontal portion h to ensure the maximum distance s covered by the disc? What is it equal to?

In order to obtain the velocity at point B, we apply the law of conservation of energy. So,

$$\begin{gathered} \text{ Loss in PE }=\text{ Gain in KE } \\ mg(H-h)=\frac{1}{2}m{v}^2 \\ v=\sqrt{\left[2g\right(H-h\left)\right]} \\ h=\frac{1}{2}g{t}^2 \end{gathered}$$

Further

$$\begin{gathered} t=\sqrt{(2h/g)} \\ s=v\times t=\sqrt{\left[2g\right(H-h\left)\right]}\times \sqrt{(2h/g)} \\ s=\sqrt{\left[4h\right(H-h\left)\right]} \end{gathered}$$                                                                     …………(i)

$\text{ For maximum value of }s,\frac{ds}{dh}=0$

$$\begin{gathered} \frac{1}{2\sqrt{\left[4h\left(H-h\right)\right]}}\times 4\left(H-2h\right)=0 \\ or \\ h=\frac{H}{2} \end{gathered}$$

Substituting $h=\frac{H}{2}$ , in equation (i), we get

$s=\sqrt{\left[4\left(HR2\right)\left(H-HR\right)\right]}=\sqrt{H^2}=H$