given :
[tex]\begin{gathered} h(x)=-16t^2+96t \\ \end{gathered}[/tex]
the parabola opens downwards. thus the maximum point is the vertex.
[tex]\begin{gathered} t(v)=-\frac{b}{2a} \\ =\frac{-96}{-32} \\ =3\sec \end{gathered}[/tex]
thus,
[tex]\begin{gathered} h(3)=-16(3)^2+96\times3 \\ =-144+288 \\ =144ft \end{gathered}[/tex]
the answer is 144ft.
