Check the picture below. When will it reach its height? well, at the vertex, so
[tex]~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&24\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&50\\ \qquad \textit{of the object}\\ h=\textit{object's height}\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ h(t)=-16t^2+24t+50\implies \\\\[-0.35em] ~\dotfill\\\\ \textit{vertex of a vertical parabola, using coefficients}[/tex]
[tex]h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+24}t\stackrel{\stackrel{c}{\downarrow }}{+50} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 24}{2(-16)}~~~~ ,~~~~ 50-\cfrac{ (24)^2}{4(-16)}\right)\implies \left( \cfrac{3}{4}~~,~~50-\cfrac{576}{64} \right)[/tex]
[tex]\left(\cfrac{3}{4}~~,~~50+9 \right)\implies \stackrel{\underset{\qquad \downarrow }{\textit{how far up it went}}}{\underset{\stackrel{\uparrow \qquad ~~}{\textit{when it reached it}}}{\left( \cfrac{3}{4}~~,~~59 \right)}}[/tex]
