Question

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 (figure shown above). 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?

Answer 1

Velocity of the body at height $h$, $v_h=\sqrt{2g(H-h)}$, horizontally (from the figure given in the problem). Time taken in falling through the distance $h$.
$t=\sqrt{\frac{2h}{g}}$ (as initial vertical component of the velocity is zero.)
Now, $s=v_{h}t=\sqrt{2g(H-h)}\times \sqrt{\frac{2h}{g}}=\sqrt{4(Hh-h^2)}$
For $s_{max}$, $\frac{d}{ds}(Hh-h^2)=0$, which yields $h=\frac{H}{2}$
Putting this value of $h$ in the expression obtained for $s$, we get,
$s_{max}=H$