Answer
$2.9\sec $
Work Step by Step
Familiarize- According to the provided statement,
$s=4.9{{t}^{2}}$
Translate- Substitute the value $s=40$ in the provided expression.
$s=4.9{{t}^{2}}$
Carry out- Substitute the value $s=40$,
$\begin{align}
& s=4.9{{t}^{2}} \\
& 40=4.9{{t}^{2}}
\end{align}$
Divide by $4.9$ on both sides,
$\begin{align}
& 40=4.9{{t}^{2}} \\
& \frac{40}{4.9}=\frac{4.9{{t}^{2}}}{4.9} \\
& \frac{40}{4.9}={{t}^{2}}
\end{align}$
Taking square roots on both sides of the equation,
$\begin{align}
& \frac{40}{4.9}={{t}^{2}} \\
& \sqrt{\frac{40}{4.9}}=\sqrt{{{t}^{2}}} \\
& 2.9\approx t
\end{align}$
Thus, Wyatt will fall for about $2.9\sec $ before the cord begins to stretch.
