so hmm check the picture below
[tex]\bf \text{initial velocity}\\
h(t) = -16t^2+v_ot+h_o \qquad \text{in feet}\\
\\ \quad \\
\begin{cases}
v_o=\textit{initial velocity of the object}\to &24\\
h_o=\textit{initial height of the object}\to &50\\
h=\textit{height of the object at "t" seconds}
\end{cases}\\\\
-----------------------------\\\\
\textit{vertex of a parabola}\\ \quad \\
y = {{ a}}x^2{{ +b}}x{{ +c}}\qquad
\left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)[/tex]
so.. the ball will reach its maximum height at [tex]\bf -\cfrac{{{ b}}}{2{{ a}}}[/tex] seconds
and will be [tex]\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}[/tex] feet above the ground
