Answer
Yes, it can resist because it can bear $360\ \text{pounds}$.
Work Step by Step
According to the given statement;
$\begin{align}
& F\propto \text{Velocit}{{\text{y}}^{2}} \\
& F\propto \text{Area} \\
& F=k\times \text{Area}\times \text{Velocit}{{\text{y}}^{2}}
\end{align}$
Substitute the values of velocity, area and force to find k using Window 1 data.
$\begin{align}
& F=k\times \text{Area}\times \text{Velocit}{{\text{y}}^{2}} \\
& 150=k\times \left( 4\times 5 \right)\times {{30}^{2}} \\
& 150=k\times \left( 20 \right)\times 900 \\
& k=\frac{150}{20\times 900}
\end{align}$
So, $k=\frac{1}{20}$
Now, $F=\frac{\text{Area}\times \text{Velocit}{{\text{y}}^{2}}}{20}$
Now, substitute the values of velocity, area and force to find k using Window 2 data.
$\begin{align}
& F=\frac{\text{Area}\times \text{Velocit}{{\text{y}}^{2}}}{20} \\
& F=\frac{\left( 3\times 4 \right)\times {{60}^{2}}}{20} \\
& F=360\ \text{pounds} \\
\end{align}$
As the hurricane force is $300\ \text{pounds}$ which less than the window 2 bearing force, it can stand against the hurricane.
