Re Drag

Consider a straight wooden fence is placed in parallel with the direction of wind flowing 20 m/s. If the fence’s Reynolds Number (Re) is 2 x 10^6 and its projected area is 0.25 square meters, what is the fence’s drag force in Newtons? Assume the density of air is 1.22 kg/m^3.

Hint
The drag force on objects immersed in a large body of flowing fluid or objects moving through a stagnant fluid:
$$$D=\frac{1}{2}\rho U^{2}C_{D}A$$$
where $$C_D$$ is the drag coefficient, $$U$$ is the flowing fluid or moving object’s velocity, $$\rho$$ is the fluid density, and $$A$$ is the projected area of blunt objects with axes perpendicular to the flow.
Hint 2
For flat plates placed parallel with the flow and have $$10^6 < Re<10^9$$ :
$$$C_D=\frac{0.031}{Re^{1/7}}$$$
where $$Re$$ is the Reynolds Number.
First, let’s find the road sign’s coefficient of drag. For flat plates placed parallel with the flow and have $$10^6 < Re<10^9$$ :
$$$C_D=\frac{0.031}{Re^{1/7}}$$$
where $$Re$$ is the Reynolds Number.
$$$C_D=\frac{0.031}{(2\cdot 10^6)^{1/7}}=\frac{0.031}{7.946}=0.0039$$$
Next, the drag force on objects immersed in a large body of flowing fluid or objects moving through a stagnant fluid:
$$$D_f=\frac{1}{2}\rho U^{2}C_{D}A$$$
where $$C_D$$ is the drag coefficient, $$U$$ is the flowing fluid or moving object’s velocity, $$\rho$$ is the fluid density, and $$A$$ is the projected area of blunt objects with axes perpendicular to the flow.
$$$D_f=\frac{1}{2}(1.22\frac{kg}{m^3}) (20m/s)^{2}(0.0039)(0.25m^2)$$$
Thus,
$$$D_f=(1.22\frac{kg}{m^3}) (400\frac{m^2}{s^{2}})(0.0039)(0.125m^2)=0.24\frac{kg\cdot m}{s^2}=0.24\:N$$$
0.24 N