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{28}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{500}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+28t+500~\hspace{8em}(\stackrel{x}{3}\qquad ,\qquad \stackrel{y}{2})[/tex]
simply means, t = 3, 3 seconds later, the rock was 2 feet up in the air.
