Two projectile thrown at 30 ° and 45 ° with the horizontal...

Time taken to reach maximum height is given by

$t = \dfrac{usin\theta}{g}$, where, $u$ is initial velocity.

Here, ${t}_{1} = {t}_{2}$

$∴ \dfrac{{u}_{1}sin{\theta}_{1}}{g} = \dfrac{{u}_{2}sin{\theta}_{2}}{g}$

$⇒ {u}_{1}\sin 30 = {u}_{2}\sin 45$

$⇒ \dfrac{{u}_{1}}{{u}_{2}} = \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{1}{2}} = \dfrac{\sqrt{2}}{1} = \sqrt{2}: 1$