A sphere of mass m, moving with velocity V, enters a hanging bag of sand and stops. If the mass of the bag is M and it is raised by height h, then the velocity of the sphere was

Consider the sand moving at a velocity of $V_s$. Often, the sphere enters using the principle of conservation of momentum.

$mV=(M+m){V_s}^2$

By the principle of energy conservation, if the system ascends to height h,

$$\begin{gathered} \frac{1}{2}\left(m_M\right){v_s}^2=(M+m)gh \\ \Rightarrow v_s=\sqrt{2gh} \\ \Rightarrow v=\left(\frac{M+m}{m}\right)\sqrt{2gh} \end{gathered}$$

Hence the correct answer is $\left(\frac{M+m}{m}\right)\sqrt{2gh}.$