Given the equation below,
[tex]y=-16t^2+64t[/tex]To find the maximum point, dy/dt = 0.
Differentiating the equation above,
[tex]\begin{gathered} y=-16t^2+64t \\ \frac{dy}{dt}=-32t+64 \end{gathered}[/tex]Where dy/dt = 0,
[tex]\begin{gathered} -32t+64=0 \\ -32t=-64 \\ t=\frac{-64}{-32}=2 \end{gathered}[/tex]Substituting for t into the equation, maximum height is'
[tex]\begin{gathered} y=-16t^2+64t \\ \text{Where t = 2} \\ y=-16(2)^2+64(2) \\ y=-16(4)+128_{} \\ y=-64+128=64 \end{gathered}[/tex]Hence, the maximum height of the football is 64 ft.
The vertex form is to be used which is given below as,
[tex]y=-16(t-2)^2+64[/tex]Where (h, k) represents the coordinate of the vertex and k is the maximum height.
