Two projectiles are projected at 30 ° and 60 ° with the horizontal...

The maximum height is given by, $H = \dfrac{{u}^{2}\sin^{2}\theta}{2g}$.

The given data is

${\theta}_{1} = 30 ° {\theta}_{2} = 60 ° {u}_{1} = {u}_{2}$

Therefore,

\dfrac{{\left({H}_{1}\right)}_\max}{{\left({H}_{2}\right)}_\max} = \dfrac{{u}_{1}^{2}\sin^{2}{\theta}_{1}}{{u}_{2}^{2}\sin^{2}{\theta}_{2}} = {\left(\dfrac{\sin 30 °}{\sin 60 °}\right)}^{2} = {\left(\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}\right)}^{2} = \dfrac{1}{3}