Two projectiles are projected with the same initial velocities at the \(15^\circ\) and \(30^\circ\) with respect to the horizontal. The ratio of their ranges is \(1:x\). The value of \(x\) is:

The horizontal range of a projectile is given by the formula \(R = \dfrac{u^2 \sin(2\theta)}{g}\).

For the first projectile, the angle of projection is \(\theta_1 = 15^\circ\).

\(R_1 = \dfrac{u^2 \sin(2 \times 15^\circ)}{g} = \dfrac{u^2 \sin(30^\circ)}{g} = \dfrac{u^2}{2g}\)

For the second projectile, the angle of projection is \(\theta_2 = 30^\circ\).

\(R_2 = \dfrac{u^2 \sin(2 \times 30^\circ)}{g} = \dfrac{u^2 \sin(60^\circ)}{g} = \dfrac{\sqrt{3}u^2}{2g}\)

The ratio of their ranges is:

\(\dfrac{R_1}{R_2} = \dfrac{\dfrac{u^2}{2g}}{\dfrac{\sqrt{3}u^2}{2g}} = \dfrac{1}{\sqrt{3}}\)

Given that the ratio is \(1:x\), we get \(x = \sqrt{3}\).

Answer: \(\sqrt{3}\)