Ratio of Ranges of Two Projectiles at Given Angles

The horizontal range $R$ of a projectile launched with an initial velocity $u$ at an angle $\theta$ with respect to the horizontal is given by the formula:

$R = \frac{u^2 \sin(2\theta)}{g}$

Given that both projectiles are launched with the same initial velocity $u$:

1. For the first projectile at an angle of projection $\theta_1 = 15^\circ$: $R_1 = \frac{u^2 \sin(2 \times 15^\circ)}{g} = \frac{u^2 \sin(30^\circ)}{g}$

Since $\sin(30^\circ) = \frac{1}{2}$: $R_1 = \frac{u^2}{2g}$

1. For the second projectile at an angle of projection $\theta_2 = 30^\circ$: $R_2 = \frac{u^2 \sin(2 \times 30^\circ)}{g} = \frac{u^2 \sin(60^\circ)}{g}$

Since $\sin(60^\circ) = \frac{\sqrt{3}}{2}$: $R_2 = \frac{\sqrt{3}u^2}{2g}$

1. Taking the ratio of their ranges $\frac{R_1}{R_2}$: $\frac{R_1}{R_2} = \frac{\sin(30^\circ)}{\sin(60^\circ)} = \frac{1/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}}$

According to the question, the ratio of their ranges is given as $1:x$, which means: $\frac{R_1}{R_2} = \frac{1}{x}$

Comparing the two expressions: $\frac{1}{x} = \frac{1}{\sqrt{3}} \implies x = \sqrt{3}$

Thus, the value of $x$ is $\sqrt{3}$, which corresponds to option B.