Answer
The gun should be aimed at an angle of $12.8^{\circ}$ above the horizontal.
Work Step by Step
We can write an expression for the initial range of the ball.
$range = \frac{v_0^2~sin(2\theta)}{g}$
$range = \frac{v_0^2~sin(2\times 30^{\circ})}{g}$
We need to find the angle when the range is half this distance.
$\frac{v_0^2~sin(2\theta)}{g} = \frac{v_0^2~sin(2\times 30^{\circ})}{2g}$
$sin(2\theta) = \frac{sin(60^{\circ})}{2}$
$sin(2\theta) = 0.433$
$\theta = \frac{arcsin(0.433)}{2}$
$\theta = 12.8^{\circ}$
The gun should be aimed at an angle of $12.8^{\circ}$ above the horizontal.
