The solution for this apparently is $\frac{1}{2}C \rho A V^{2}$ where $C$ is the drag coefficient, $A$ is the area and $\rho$ and $V$ are the wind density and velocity. Let's do it without the drag coefficient.
Assume that the wind hits the panel normally with speed $V$ and is then stopped (or escapes along the panel). The change in momentum of the wind per unit time is
\begin{align}
F&=\frac{d}{dt}M_{wind}V\\
&=\frac{d}{dt}(\rho A h) V\\
&=\rho A V^{2},
\end{align}
where $h$ is some height above the panel. Thus, the force on the panel is $\rho A V^{2}$. What happened to the factor of $1/2$?
Best Answer
That's not the best of assumptions. The air will flow around the plate. Only in the center of the plate will it hit the panel normally and stop. Your analysis does however capture some of the key dynamics in that the drag force is proportional to $\rho A V^2$.
Some shapes will do a much better job directing the flow around the object than others. Consider the drag on a flat plate, a sphere, and a nicely shaped airfoil, all with the same cross sectional area. The flat plate is going to do a rather lousy job redirecting the air flow and will experience a large drag. The sphere will do a good deal better than a flat plate and suffer less drag. The airfoil will do an even better job and suffer even less drag that the sphere. That the shape of an object has a marked impact on the drag force is what the coefficient of drag tries to capture.
Update: Why a factor of 1/2?
A few simplifying assumptions lead directly to that factor of 1/2. I'll assume a somewhat slowly moving fluid. Not so slowly as to result in Stokes drag, but not so quickly as to result in turbulence. That means a Reynolds number between 10 and 2000 or so. With these assumptions, the flow will be dominated by convective acceleration but will still be steady.
Consider a packet of fluid moving horizontally with velocity $v$, density $\rho$, and pressure $p$. Given the assumptions, the conservation laws lead to a simplified form of Bernoulli's equation, $\frac {d}{dx} \left( \rho \frac {v^2} 2 + p \right) = 0$. You can find many derivations of Bernoulli's equation on the internet. The term $\rho \frac {v^2} 2$ is the dynamic pressure that results from the flow. Naively, this leads to the force on an object being the product of the dynamic pressure and the cross section area: $F = \rho A \frac {v^2} 2$.
This is not quite the case. As noted earlier, different objects with the same cross section have markedly different drag forces. The coefficient of drag is a unitless parameter that accounts for these variations. Also note that the coefficient of drag is not a constant. It is a function of both shape and velocity.