so hmmm check the picture below
the leading term's coefficient for that quadratic is negative, meaning is opening downwards
so.. hmm where's the vertex anyway? well
[tex]\bf \textit{vertex of a parabola}\\ \quad \\
h=-16t^2+128t\\\\\\
\begin{array}{lccclll}
h=&-16t^2&+128t&+0\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)[/tex]
so,the cannonball reaches a maximum height of [tex]\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}[/tex]
