Time taken to reach maximum height $$ \begin{aligned} & \mathrm{t}=\frac{\mathrm{u}_{\sin \theta}}{\mathrm{g}} \quad \therefore \frac{\mathrm{u}_1 \sin \theta_1}{\mathrm{~g}}=\frac{\mathrm{u}_2 \sin \theta_2}{\mathrm{~g}} \\ & \Rightarrow \mathrm{u}_1 \sin 30=\mathrm{u}_2 \sin 45 \Rightarrow \frac{\mathrm{u}_1}{\mathrm{u}_2}=\frac{1 / \sqrt{2}}{1 / 2}=\frac{\sqrt{2}}{1} \end{aligned} $$