1. Use the formula for time to reach maximum height

For a projectile thrown with initial speed $u$ at angle $\theta$, the vertical component is:

$u_y = u\sin\theta$

Time to reach maximum height is:

$t = \frac{u\sin\theta}{g}$

2. Apply the condition given in the question

Let the initial velocities of the two projectiles be $u_1$ and $u_2$.

• First projectile: angle $30^\circ$
• Second projectile: angle $45^\circ$

Since both reach maximum height in the same time,

$\frac{u_1\sin 30^\circ}{g} = \frac{u_2\sin 45^\circ}{g}$

Cancelling $g$,

$u_1\sin 30^\circ = u_2\sin 45^\circ$

3. Substitute the values of sines

$\sin 30^\circ = \frac{1}{2}, \qquad \sin 45^\circ = \frac{1}{\sqrt{2}}$

So,

$u_1\left(\frac{1}{2}\right) = u_2\left(\frac{1}{\sqrt{2}}\right)$

$\frac{u_1}{u_2} = \frac{2}{\sqrt{2}} = \sqrt{2}$

Hence,

$u_1 : u_2 = \sqrt{2} : 1$

4. Match with the options

This corresponds to Option C.

Final Answer

$\boxed{\sqrt{2}:1}$