Two projectiles thrown at $30^{\circ}$ and $45^{\circ}$ with the horizontal respectively, reach the maximum height in same time. The ratio of their initial velocities is :

<p>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.</p> <p>The formula to calculate the time to reach the maximum height is given by:</p> <p> <p>$t = \frac{u \sin \theta}{g}$</p> </p> <p>where:</p> <ul> <li>$t$ is the time to reach maximum height,</li> <li>$u$ is the initial velocity,</li> <li>$\theta$ is the angle of projection, and</li> <li>$g$ is the acceleration due to gravity.</li> </ul> <p>Given that the projectiles reach the maximum height in the same time, we can set up the following equation:</p> <p> <p>$\frac{u_1 \sin 30^\circ}{g} = \frac{u_2 \sin 45^\circ}{g}$</p> </p> <p>Since $\sin 30^\circ = \frac{1}{2}$ and $\sin 45^\circ = \frac{\sqrt{2}}{2}$, the equation simplifies to:</p> <p> <p>$\frac{u_1 \cdot \frac{1}{2}}{g} = \frac{u_2 \cdot \frac{\sqrt{2}}{2}}{g}$</p> </p> <p>Canceling out the common terms (i.e., $g$ and $\frac{1}{2}$), we get:</p> <p> <p>$u_1 = u_2 \sqrt{2}$</p> </p> <p>Hence, the ratio of their initial velocities is:</p> <p> <p>$\frac{u_1}{u_2} = \sqrt{2}$</p> </p> <p>Therefore, the correct answer is Option C: $\sqrt{2}:1$.</p>