A ball is thrown vertically upwards from the top of a tower. Velocity at a point 'h' m vertically below the point of projection is twice the downward velocity at a point 'h' m vertically above the point of projection. The maximum height reached by the ball above the top of
the tower is

Let v represent the speed and K represent the kinetic energy at height h above the tower. The velocity will therefore be 2v at height h below the tower. At this time, kinetic energy will be 4K because $K\propto v^2$. So, through energy conservation,

$$\begin{gathered} K+mgh=4K-mgh \\ K=\frac{2mgh}{3} \end{gathered}$$

Also, let H be the maximum height that can be reached over the tower. then, once more using conserving energy,

$$\begin{gathered} K+mgh=\frac{2mgh}{3}+mgh=mgH \\ \Rightarrow H=\frac{5h}{3} \end{gathered}$$

Hence the correct answer is $\frac{5h}{3}.$