Two projectiles are projected with the same initial velocities at...

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: $\sqrt{3}$