Answer
$\text{11}\text{.4sec}$
Work Step by Step
$\left( s=2063\text{ft} \right)$
Put the value of $s$ in the expression,
$\begin{align}
& s=16{{t}^{2}} \\
& 2063=16{{t}^{2}} \\
& \frac{2063}{16}={{t}^{2}}
\end{align}$
And,
$\begin{align}
& \frac{2063}{16}={{t}^{2}} \\
& 128.9375={{t}^{2}}
\end{align}$
Use the principle of square roots,
$\begin{align}
& {{t}^{2}}=128.9375 \\
& t=\sqrt{128.9375} \\
& \approx \pm 11.35
\end{align}$
We only consider the positive values of time.
Therefore, the approximate value of $t$ is,
$t\approx 11.4\text{sec}$
