Question

A ball of mass m falls from height h under
gravity. Let v and k denote the potential and
kinetic energy of the ball at height x. When x
2h/3, the ratio k/v is :​

Answer 1

$\text{Potential Energy at} \frac{2h}{3}=\boxed{ \frac{2mgh}{3}}\\ </p><p>\text{Kinetic Energy at} \frac{2h}{3}= \frac{mv^2}{2}$

Since the body falls through h/3, its velocity at that height is $v=\sqrt{\frac{2gh}{3}}\\</p><p>\mathrm{ Hence, \: Kinetic \: energy \: is }\\</p><p>K.E.= \frac{m×2hg}{2×3}\\</p><p>K.E.=\boxed{\frac{mgh}{3}}\\</p><p>\mathrm{Thus, \: ratio} \: =\textcolor{red}{\frac{2}{1}}$