check the picture below.
[tex]\bf h(t)=40t-16t^2\implies h(t)=-16t^2+40t+0
\\\\\\
\begin{array}{llll}
\qquad \textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o
\end{array}
\quad
\begin{cases}
v_o=\textit{initial velocity of the object}\\
\qquad 40\\
h_o=\textit{initial height of the object}\\
\qquad 0\\
h=\textit{height of the object at "t" seconds}
\end{cases}[/tex]
[tex]\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{llll}
y = &{{ -16}}x^2&{{ +40}}x&{{ +0}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
\\\\\\
\left( \stackrel{\textit{it took this long}}{-\cfrac{40}{2(-16)}}~~,~~\stackrel{\textit{to get this high}}{0-\cfrac{40^2}{4(-16)}} \right)\qquad \stackrel{\textit{maximum height}}{0-\cfrac{40^2}{4(-16)}}[/tex]
