check the picture below, it hits the ground at y = 0
thus
[tex]\bf \qquad \textit{initial velocity}\\\\
\begin{array}{llll}
\qquad \textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o
\end{array}
\quad
\begin{cases}
v_o=\textit{initial velocity of the object}\\
h_o=\textit{initial height of the object}\\
h=\textit{height of the object at "t" seconds}\\
----------\\
v_o=208\\
h_o=0
\end{cases}[/tex]
[tex]\bf h(t)=-16t^2+208t+0\implies h(t)=-16t^2+208t\impliedby
\begin{array}{llll}
\textit{now let's}\\
make\\ h(t)=0
\end{array}
\\\\\\
0=-16t^2+208t\implies 0=t(208-16t)
\\\\\\
\begin{cases}
0=t\\
----------\\
0=208-16t\\
\qquad 16t=208\\
\qquad t=\frac{208}{16}
\end{cases}[/tex]
and I'm sure you know how much that is
it hits it at two points, as you notice in the picture, at 0 seconds, when it's first hit and goes up, and when it falls down.
