Answer
(a) $63.4^{\circ}$
(b) No.
(c) $\frac{\pi}{2}$ or $90^{\circ}$
Work Step by Step
(a) With $M(\theta)=2000ft, v_0=400ft/s$, the equation becomes $2000=\frac{400^2sin^2\theta}{64}$ which leads to $sin^2\theta=0.8$ and $sin\theta=\pm0.8944$. As $0\leq\theta\leq\frac{\pi}{2}$, we have $\theta=63.4^{\circ}$
(b) Since $sin^2\theta\leq1$, we have $M(\theta)\leq\frac{400^2}{64}=2500ft$, thus it is not possible for the projectile to reach a height of 3000 ft.
(c) To reach a maximum height of 2500ft, it is required that $\theta=\frac{\pi}{2}$ or $90^{\circ}$
