check the picture below.
so, the maximum height is reached at the parabola's vertex, where x = how many seconds it took, y = how hight it went.
[tex]\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
h = &{{ -16}}t^2&{{ +36}}t&{{ +10}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)[/tex]
[tex]\bf \left(-\cfrac{36^2}{2(-16)}~~,~~10-\cfrac{36^2}{4(-16)} \right)\implies \left( \cfrac{1296}{32}~~,~~10+\cfrac{81}{4} \right)
\\\\\\
\left( \cfrac{81}{2}~~,~~\cfrac{121}{4} \right)\implies \left(\stackrel{\textit{how many seconds}}{40\frac{1}{2}}~~,~~\stackrel{\textit{how many feet up}}{30\frac{1}{4}} \right)[/tex]
