A ball P is projected at an angle of $60^\circ$ with the vertical with certain initial speed. Another ball Q of the same mass as that of ball P is projected vertically upwards with the same initial speed as that of P. At the highest point, the ratio of potential energy of ball P to that of ball Q is
$(\sin 30^\circ = 0.5)$

Let the initial speed of both balls be $u$ and their mass be $m$.

For ball P, the angle of projection with the vertical is $60^\circ$, so the angle with the horizontal is $\theta_P = 90^\circ - 60^\circ = 30^\circ$.

The maximum height reached by ball P is $H_P = \dfrac{u^2 \sin^2 30^\circ}{2g} = \dfrac{u^2 (1/2)^2}{2g} = \dfrac{u^2}{8g}$.

The potential energy of ball P at the highest point is $U_P = mgH_P = mg \left(\dfrac{u^2}{8g}\right) = \dfrac{mu^2}{8}$.

For ball Q, it is projected vertically upwards, so the angle of projection with the horizontal is $\theta_Q = 90^\circ$.

The maximum height reached by ball Q is $H_Q = \dfrac{u^2 \sin^2 90^\circ}{2g} = \dfrac{u^2}{2g}$.

The potential energy of ball Q at the highest point is $U_Q = mgH_Q = mg \left(\dfrac{u^2}{2g}\right) = \dfrac{mu^2}{2}$.

The ratio of their potential energies at the highest point is $\dfrac{U_P}{U_Q} = \dfrac{\dfrac{mu^2}{8}}{\dfrac{mu^2}{2}} = \dfrac{2}{8} = \dfrac{1}{4}$.

Answer: $1 : 4$