[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{24}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{50}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}
\\\\\\
h(t)=-16t^2+24t+50[/tex]
now, check the picture below, it reaches a maximum point at the vertex, where the seconds is the x-coordinate and the height is the y-coordinate.
so, let's use the coefficients to get the vertex out of that quadratic then,
[tex]\bf \textit{ vertex of a vertical parabola, using coefficients}\\\\
\begin{array}{lccclll}
h(t) = &{{ -16}}t^2&{{ +24}}t&{{ +50}}\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)[/tex]
[tex]\bf \left(-\cfrac{24}{2(-16)}~~,~~50-\cfrac{24^2}{4(-16)} \right)\implies \left( \cfrac{3}{4}~~,~~50- \cfrac{576}{-64}\right)
\\\\\\
\left( \cfrac{3}{4}~~,~~50+ 9\right)\implies \left(\stackrel{\textit{seconds}}{\frac{3}{4}}~~,~~\stackrel{feet}{59} \right)[/tex]
