A stone is projected at angle 30 ° to the horizontal. The ratio of...

Let stone is projected with initial velocity $v$.

Kinetic energy of the stone at point of projection is $K{E}_{1} = \dfrac{1}{2}m{v}^{2}$.

Now, velocity of stone at maximum height is $= vcos30 °$, so kinetic energy at that point is $K{E}_{2} = \dfrac{1}{2}m{\left(vcos30 °\right)}^{2}$

Hence, the ratio $\dfrac{K{E}_{1}}{K{E}_{2}} = \dfrac{\dfrac{1}{2}m{v}^{2}}{\dfrac{1}{2}m{\left(vcos30 °\right)}^{2}} = \dfrac{4}{3}$