Step 1: Recall the range formula.
For a projectile launched with speed $u$ at angle $\theta$, the horizontal range is $R = \dfrac{u^{2} \sin 2\theta}{g}$. The range depends on the angle only through $\sin 2\theta$.
Step 2: Find the maximum range.
$\sin 2\theta$ is largest when $2\theta = 90^\circ$, that is $\theta = 45^\circ$. Then $\sin 90^\circ = 1$, so $R_{max} = \dfrac{u^{2}}{g}$.
Step 3: Find the range at 30 degrees.
At $\theta = 30^\circ$ we use $2\theta = 60^\circ$. So $R_{30} = \dfrac{u^{2} \sin 60^\circ}{g}$.
Step 4: Put in the value of sin 60.
Since $\sin 60^\circ = \dfrac{\sqrt{3}}{2} \approx 0.866$, we get $R_{30} = 0.866\, R_{max}$.
Step 5: Find the drop in range.
The range falls from $R_{max}$ to $0.866\, R_{max}$. The decrease is $R_{max} - 0.866 R_{max} = 0.134\, R_{max}$.
Step 6: Turn it into a percentage.
Divide the drop by the maximum range and multiply by 100: $\dfrac{0.134\, R_{max}}{R_{max}} \times 100 = 13.4\%$. \[ \boxed{13.4} \]