so hmmm check the graph below
now.. to find the vertex of any quadratic
[tex]\bf \textit{vertex of a parabola}\\ \quad \\
\begin{array}{lccclll}
h(t)=&-16t^2&+16t&+480\\
&\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... how long did it take? well, it took [tex]\bf -\cfrac{{{ b}}}{2{{ a}}}[/tex]
what was the highest point? well, it was [tex]\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}[/tex]
when did he hit the water? well, at y = 0
that is [tex]h(t) = -16t^2 + 16t + 480 \implies 0 = -16t^2 + 16t + 480,[/tex]
solve for "t"