check the picture below.
so the ball reaches a maximum height at its vertex,
[tex] \bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
h=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+32}t\stackrel{\stackrel{c}{\downarrow }}{+6}
\qquad \qquad
\left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left(-\cfrac{32}{2(-16)}~~,~~6-\cfrac{32^2}{4(-16)} \right)\implies \left(\cfrac{32}{32}~~,~~6-\cfrac{1024}{-64} \right)
\\\\\\
(1~~,~~6+16)\implies (\stackrel{seconds}{1}~,~\stackrel{\stackrel{feet}{high}}{22}) [/tex]