Physics - Kinematics Question with Solution | TestHub

Time taken to reach maximum height is given by 

$t=\frac{u\sin \theta }{g}$, where, $u$ is initial velocity.

Here, $t_1=t_2$

$\therefore \frac{u_1\sin {\theta }_1}{g}=\frac{u_2\sin {\theta }_2}{g}$

$\Rightarrow u_1\sin 30=u_2\sin 45$

$\Rightarrow \frac{u_1}{u_2}=\frac{{\displaystyle \frac{1}{\sqrt{2}}}}{{\displaystyle \frac{1}{2}}}=\frac{\sqrt{2}}{1}=\sqrt{2}:1$