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. What must be the height of the the vertical portion $h$ to ensure that a maximum distance $s$ is covered by the disc, when it leaves at point B? What is the value of $s$?

• A. $\frac{H}{2}$, $H$
  
• B. $\frac{H}{2}$, $\frac{H}{2}$
• C. $2H$, $\frac{H}{2}$
• D. $\frac{H}{3}$, $\frac{H}{2}$

Answer 1

The correct option is A $\frac{H}{2}$, $H$

In order to obtain the velocity at point. B, we apply the law of conservation of energy. So,
Loss in PE $=$Gain in KE
$mg(H-h)=\frac{1}{2}m{v}^2$

$\therefore v=\sqrt{\left[2g(H-h)\right]}$

Further $h=\frac{1}{2}g{t}^2$

$\therefore t=\sqrt{\frac{2h}{g}}$
Now, $s=v\times t=\sqrt{\left[2g(H-h)\right]}\times \sqrt{\frac{2h}{g}}$

or $s=\sqrt{\left[4h(H-h)\right]}$
For maximum value of s: $\frac{ds}{dh}=0$

Alternatively, sum of h and (H-h) is H but product has to be maximized.
This happens, when $h=H-h$
$\therefore \frac{1}{2\sqrt{\left[4h(H-h)\right]}}\times 4(H-2h)=0$
or $h=\frac{H}{2}$
Substituting $h=\frac{H}{2}$, in Eq. (i), we get
$s=\sqrt{\left[4\left(H/2\right)(H-H/2)\right]}=\sqrt{H^2}=H$