Question

A small sphere starts sliding down on a frictionless hill from point $\text{A}$ as shown. What must be the height of the horizontal portion $h$, to ensure that the sphere covers the maximum horizontal distance $s$ ?

• A. $h=\frac{H}{3}$
• B. $h=\frac{H}{4}$
• C. $h=\frac{H}{2}$
• D. $h=\frac{2H}{3}$

Answer 1

The correct option is C $h=\frac{H}{2}$

Since, there are no external and non-conservative forces acting on the sphere, so the total energy of the sphere remains constant from $\text{A}$ to $\text{B}$.

$\therefore$ Loss in $PE$ = Gain in $KE$

$mgH-mgh=\frac{1}{2}m{v}^2-0$

$\therefore v=\sqrt{2g(H-h)}$

From point $\text{B}$ onwards, the sphere performs horizontal projectile motion,

$\therefore \text{Range}\left(R\right)=s=v\times \text{time of flight}\left(T\right)$

$\Rightarrow s=\sqrt{2g(H-h)}\times \sqrt{\frac{2h}{g}}$

$\therefore s=\sqrt{4h(H-h)}$

For $s$ to be maximum,

$\frac{ds}{dh}=0$

$\frac{ds}{dh}=\frac{1}{2\sqrt{4h(H-h)}}\times 4(H-2h)=0$

$\Rightarrow \frac{H-2h}{\sqrt{h(H-h)}}=0$

From the above equation, we get

$H-2h=0$ and $H-h\ne 0$

$\therefore h=\frac{H}{2}$

Hence, option $\left(C\right)$ is correct.