Given the equation below,
![y=-16t^2+64t](https://img.qammunity.org/2023/formulas/mathematics/college/dwdsdmp4hkr0uns2p98t7ml1k9ycr42qr0.png)
To find the maximum point, dy/dt = 0.
Differentiating the equation above,
![\begin{gathered} y=-16t^2+64t \\ (dy)/(dt)=-32t+64 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/i0yg2zpdhu6xmlg7hvk0rq3008wow8g422.png)
Where dy/dt = 0,
![\begin{gathered} -32t+64=0 \\ -32t=-64 \\ t=(-64)/(-32)=2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/f0z1z1e23msdc94hyt32rsgq61lhct19yk.png)
Substituting for t into the equation, maximum height is'
![\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}](https://img.qammunity.org/2023/formulas/mathematics/high-school/oc4mb97u6toyaz1cc69brfxqn6oeand39u.png)
Hence, the maximum height of the football is 64 ft.
The vertex form is to be used which is given below as,
![y=-16(t-2)^2+64](https://img.qammunity.org/2023/formulas/mathematics/high-school/6gdl4x833qwpfqwdw9ztf7nj346qpeuqei.png)
Where (h, k) represents the coordinate of the vertex and k is the maximum height.