Answer
Result \(\alpha \approx 24.3^\circ\), \(\alpha \approx 65.7^\circ\)
Work Step by Step
Consider the angle as \(\alpha\) and the initial velocity \(v_0\), and the distance of three fourths the maximum range of the shell. Since the range given by \[ R = \frac{g v_0^2 \sin(2\alpha)}{g} = v_0^2 \sin(2\alpha) \] and the maximum range is \[ R_{\text{max}} = \frac{v_0^2}{g} \] Then \[ \frac{ v_0^2 \sin2 \alpha}{g} = \frac{3}{4} \frac{v_0^2}{g} \] \[ \sin(2\alpha) = \frac{3}{4} \] \[ \alpha = \frac{1}{2} \sin^{-1}\left(\frac{3}{4}\right) \approx 24.3^\circ \] or \[ \alpha = 180^\circ - 24.3^\circ \approx 65.7^\circ \] Result \(\alpha \approx 24.3^\circ\), \(\alpha \approx 65.7^\circ\)
