If the two projectile are projected with the same speed and the angles of projection for two projectiles are 30° and 60°…

The maximum height of the projectile is given by

\(\mathrm{H}=\frac{{\mathrm{u}}^{2}{\sin }^{2}\theta }{2\mathrm{g}}\\ \mathrm{So},\\ \frac{{\mathrm{H}}_{1}}{{\mathrm{H}}_{2}}=\frac{{\sin }^{2}{\theta }_{1}}{{\sin }^{2}{\theta }_{2}}=\frac{{\sin }^{2}60^\circ }{{\sin }^{2}30^\circ }\\ \frac{{\mathrm{H}}_{1}}{{\mathrm{H}}_{2}}=\sqrt{3}:1\)