check the picture below. So the rocket reaches its height at its vertex.
the rocket is being launched from the ground, and therefor its initial height is 0, thus
[tex] \bf ~~~~~~\textit{initial velocity}
\\\\
\begin{array}{llll}
~~~~~~\textit{in feet}
\\\\
h(t) = -16t^2+v_ot+h_o
\end{array}
\quad
\begin{cases}
v_o=\stackrel{352}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{0}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+352t+0 [/tex]
so we can get the vertex coordinates by using its coefficients, how many "t" seconds will it take? well, that'd be the x-coordinate of its vertex.
[tex] \bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+352}t\stackrel{\stackrel{c}{\downarrow }}{+0}
\qquad \qquad
\left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left(-\cfrac{352}{2(-16)}~~,~~\qquad \right)\implies \left(\cfrac{352}{32}~~,~~\qquad \right)\implies (\stackrel{seconds}{11}~~,\qquad ) [/tex]
