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 $B,\Delta U+\Delta K=0$
$\Rightarrow \left(U_B-U_A\right)+\left(K_B-K_A\right)=0$
$-mg(H-h)+\left(\frac{1}{2}m{v}_B^2-0\right)=0$
$\Rightarrow \frac{1}{2}m{v}_B^2=mg(H-h)$ or $v_B=\sqrt{2g(H-h)}$
The time to fall $t=\sqrt{\frac{2h}{g}}$ Range $=v_B\cdot t=\sqrt{2g(H-h)}\sqrt{\frac{2h}{g}}$
Range $=2\sqrt{(H-h)h}$
Range will be maximum, when (Range) ${}^2$ is maximum $f(h)=(H-h)h=Hh-h^2$
$f^{\prime }(h)=H-2h=0\Rightarrow h=\frac{H}{2}$
$f^{\prime }(h)=-2<0\Rightarrow$ Maxima