Question

A ball of mass m is dropped from a cliff of height $H$ . The ratio of its kinetic energy to the potential energy when it is fallen through a height $3/4H$ is

• A. $3:4$
• B. $4:3$
• C. $1:3$
• D. $3:1$

Answer 1

Answer: (C)

Total mechanical energy at height, $H$
$E_H=mgH$
Let $v_h$ be velocity of the ball at height $h$ $\left(=\frac{3}{4}H\right)$ .
$\therefore$ Total mechanical energy at height $h$ ,
$E_h=mgh+\frac{1}{2}m{v}_h^2$
According to law of conservation of mechanical energy,
$E_H=E_h;mgH=mgh+\frac{1}{2}m{v}_h^2$
$v_h{}^2=2g\left(H-h\right)$
Required ratio of kinetic energy to potential energy at height h is
$\frac{K_h}{V_h}=\frac{\frac{1}{2}m{v}_h^2}{mgh}$
$=\frac{\frac{1}{2}m2g\left(H-h\right)}{mgh}=\left(\frac{H}{h}-1\right)=\frac{1}{3}$