[tex]~~~~~~\textit{initial velocity in feet} \\\\ h(t) = -16t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}&96\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}&160\\ \qquad \textit{of the object}\\ h=\textit{object's height}&h\\ \qquad \textit{at "t" seconds} \end{cases} \\\\\\ h(t)=-16t^2+96t+160[/tex]
Check the picture below.
[tex]\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-16}x^2\stackrel{\stackrel{b}{\downarrow }}{+96}x\stackrel{\stackrel{c}{\downarrow }}{+160} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right) \\\\\\ \left(-\cfrac{ 96}{2(-16)}~~~~ ,~~~~ 160-\cfrac{ (96)^2}{4(-16)}\right) \implies \left( - \cfrac{ 96 }{ -32 }~~,~~160 - \cfrac{ 9216 }{ -64 } \right)[/tex]
[tex](3~~,~~160 + 144)\implies \underset{~\hfill feet}{\stackrel{seconds~\hfill }{(\stackrel{\downarrow }{3}~~,~~\underset{\uparrow }{304})}}[/tex]
