Question

A small body of mass m slides down from the top A of a. smooth hill of height H with initial velocity zero. The hill becomes horizontal at the edge B whose height is h from the ground. Assume that the body falls at maximum distance away from the foot of the hill. Then the value of h is

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

Answer 1

Answer: (B)

Between points $A$ and $B, \Delta U+\Delta K=0$
$\Rightarrow \left(U_{B}-U_{A}\right)+\left(K_{B}-K_{A}\right)=0$
$-m g(H-h)+\left(\frac{1}{2} m v_{B}^{2}-0\right)=0$
$\Rightarrow \frac{1}{2} m v_{B}^{2}=m g(H-h)$ or $v_{B}=\sqrt{2 g(H-h)}$
The time to fall $t=\sqrt{\frac{2 h}{g}}$ Range $=v_{B} \cdot t=\sqrt{2 g(H-h)} \sqrt{\frac{2 h}{g}}$
Range $=2 \sqrt{(H-h) h}$
Range will be maximum, when (Range) $^{2}$ is maximum $f(h)=(H-h) h=H h-h^{2}$
$f'(h)=H-2 h=0 \Rightarrow h=\frac{H}{2}$
$f'(h)=-2<0 \Rightarrow $ Maxima