Check the picture below.
so if we just find its vertex, we know how many feet it went up by its y-coordinate.
[tex]\bf h=64t-32t^2\implies h=-32t^2+64t+0 \\\\[-0.35em] ~\dotfill[/tex]
[tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h=\stackrel{\stackrel{a}{\downarrow }}{-32}t^2\stackrel{\stackrel{b}{\downarrow }}{+64}t\stackrel{\stackrel{c}{\downarrow }}{+0} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{64}{2(-32)}~~,~~0-\cfrac{64^2}{4(-32)} \right)\implies \left( \stackrel{\stackrel{\textit{how many}}{\textit{seconds}}}{1}~~,~~\stackrel{\stackrel{\textit{how many feet}}{\textit{it went up}}}{32} \right)[/tex]
