A ball of mass $2 \,kg$ and another of mass $4\, kg$ are dropped together from a $60$ feet tall building. After a fall of $30$ feet each towards earth, their respective kinetic energies will be in the ratio of
Here $V _{1}= V _{2}=\sqrt{2 gh }$ $\frac{ KE _{1}}{ KE _{2}}=\frac{\frac{1}{2} m _{1} v _{1}^{2}}{\frac{1}{2} m _{2} v _{2}^{2}}$ $=\frac{ m _{1}}{ m _{2}}=\frac{2}{4}=\frac{1}{2}$