QJEEM22Ph77b9: JEE Main 2022 Physics - Class 11 JEEM22Ph77b9 Solution: Two projectiles thrown at and with the horizontal respectively, …

To solve this problem, we need to understand the relationship between the angles of projection, the initial velocities, and the time taken to reach maximum height for each projectile.

The formula to calculate the time to reach the maximum height is given by:

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

where:

• $t$ is the time to reach maximum height,
• $u$ is the initial velocity,
• $\theta$ is the angle of projection, and
• $g$ is the acceleration due to gravity.

Given that the projectiles reach the maximum height in the same time, we can set up the following equation:

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

Since $\sin 30^\circ = \frac{1}{2}$ and $\sin 45^\circ = \frac{\sqrt{2}}{2}$ , the equation simplifies to:

$\frac{u_1 \cdot \frac{1}{2}}{g} = \frac{u_2 \cdot \frac{\sqrt{2}}{2}}{g}$

Canceling out the common terms (i.e., $g$ and $\frac{1}{2}$ ), we get:

$u_1 = u_2 \sqrt{2}$

Hence, the ratio of their initial velocities is:

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

Therefore, the correct answer is Option C: $\sqrt{2}:1$ .