Given,
The given parametric equation is,
[tex]\begin{gathered} x=(v_0\text{ }cos\text{ }\theta)t \\ y=h+(v_0\text{ }\sin \text{ }\theta)t-16t^2 \end{gathered}[/tex]Subsituting the value of
[tex]\begin{gathered} \theta=60^{\circ}_{} \\ v_0=93\text{ feet per second} \end{gathered}[/tex]The equation becomes,
[tex]\begin{gathered} x=(93cos\text{ }60)t \\ x=(93\times\frac{1}{2})t \\ x=46.5t \\ y=h+(46.5\sqrt[]{3}_{})t-16t^2 \end{gathered}[/tex]Taking t =1,
then,
x = 46.5,
y = 111.04.
Similarly taking t = 2,
Then x = 93 and y = 190. 081
Hence, graph c is correct option