## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The gun should be aimed at an angle of $12.8^{\circ}$ above the horizontal.
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.