Check the picture below, so the ball's path is pretty much like so, and it reaches its hightest at its vertex, so
[tex]~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&80\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&0\\ \qquad \textit{of the object}\\ h=\textit{object's height}&h\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ f(t)=80t-16t^2\implies f(t)=-16t^2+80t+0 \\\\[-0.35em] ~\dotfill[/tex]
[tex]\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-16}x^2\stackrel{\stackrel{b}{\downarrow }}{+80}x\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)[/tex]
[tex]\left(-\cfrac{ 80}{2(-16)}~~~~ ,~~~~ 0-\cfrac{ (80)^2}{4(-16)}\right) \implies \left( - \cfrac{ 80 }{ -32 }~~,~~0 - \cfrac{ 6400 }{ -64 } \right) \\\\\\ \left( \cfrac{5}{2}~~,~~100 \right)\implies \underset{~\hfill feet ~~ }{\stackrel{seconds\qquad }{\left( 2\frac{1}{2}~~,~~100 \right)}}[/tex]
