A stone projected vertically upwards from the ground reaches a maximum height h. When it is at a height $\frac{h}{4}$  the ratio of kinetic and potential energies is

We know that energy is always conserved
Potential energy  $P.E=mgh \Rightarrow P.E=mg\frac{h}{u}$
 
Kinetic energy  $K.E=\frac{1}{2}m{v}^2 v^2-u^2=2gh$
 
For kinetic energy
 $$\begin{gathered} h=\frac{3h}{4} v^2=2g\times \frac{3h}{4}=\frac{3gh}{2} \therefore K.E=\frac{1}{2}m\times \frac{3gh}{2}=\frac{3mgh}{4} \\ \therefore \frac{KE}{PE}=\frac{\frac{3mgh}{4}}{\frac{mgh}{4}}=3:1 \end{gathered}$$