JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Work, Power & Energy

A hollow sphere has moment of inertia $I=\frac{2}{3}m{R}^2$, so when it rolls without slipping its total kinetic energy is $\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}m{v}^2+\frac{1}{2}\cdot \frac{2}{3}m{R}^2\cdot \frac{v^2}{R^2}=\frac{5}{6}m{v}^2$. At the highest point the ball is momentarily at rest and all this energy has become gravitational potential energy, so $\frac{5}{6}m{v}^2=mgh$. Hence $h=\frac{5v^2}{6g}=\frac{5(3{)}^2}{6(10)}=0.75\text{m}$. The maximum height reached above the initial position is $75\text{cm}$.