Solution:
The range of the projectile is the displacement in the horizontal direction. It is given by the equation;
[tex]\begin{gathered} r=\frac{1}{32}v^2_o\sin (2\theta) \\ \text{Where;} \\ v_o=\text{initial velocity} \end{gathered}[/tex]
Thus, given;
[tex]\begin{gathered} r=327ft \\ v_o=111\frac{ft}{s} \end{gathered}[/tex]
We have;
[tex]\begin{gathered} 327=\frac{1}{32}(111)^2\sin (2\theta) \\ 327=\frac{1}{32}(12321)\sin (2\theta) \\ 327=385.03\sin (2\theta) \\ \sin (2\theta)=\frac{327}{385.03} \\ \sin (2\theta)=0.8493 \\ 2\theta=\sin ^{-1}(0.8493) \end{gathered}[/tex][tex]\begin{gathered} 2\theta=58.1336 \\ \theta=\frac{58.1336}{2} \\ \theta=29.07^o \end{gathered}[/tex]
