check the picture below.
[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{48}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{6}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+48t+6[/tex]
so the ball will reach its highest at the "vertex", let's check then.
[tex]\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+48}t\stackrel{\stackrel{c}{\downarrow }}{+6}
\qquad \qquad
\left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left( -\cfrac{48}{2(-16)}~~,~~6-\cfrac{48^2}{4(-16)} \right)\implies \left(\cfrac{3}{2}~~,~~6+36 \right)
\\\\\\
\left( \stackrel{\textit{how many seconds}}{1\frac{1}{2}}~~,~~ \stackrel{\textit{maximum height}}{42} \right)[/tex]
