Consider a basketball hoop’s backboard is placed in parallel with the direction of wind such that it experiences a 0.5 N drag force. If the backboard’s Reynolds Number (Re) is 4 x 10^4 and its projected area is 0.75 square meters, what is the wind speed in m/s? 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 $$10^4 < Re<5\cdot10^5$$ : $$C_D=\frac{1.33}{\sqrt{Re}}$$$
where $$Re$$ is the Reynolds Number.
First, let’s find the backboard’s coefficient of drag. For flat plates placed parallel with the flow and have $$10^4 < Re<5\cdot10^5$$ :
$$C_D=\frac{1.33}{\sqrt{Re}}$$$where $$Re$$ is the Reynolds Number. $$C_D=\frac{1.33}{\sqrt{4\cdot 10^4}}=\frac{1.33}{200}=0.00665$$$
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. $$0.5N=\frac{1}{2}(1.22\frac{kg}{m^3}) U^{2}(0.00665)(0.75m^2)$$$
Solving for velocity:
$$U^{2}=\frac{1N}{0.00608\frac{kg}{m}}=\frac{1\frac{kg\cdot m}{s^2}}{0.00608\frac{kg}{m}}=164.34\:m^2/s^2$$$Thus, $$U=\sqrt{164.34m^2/s^2}=12.8\:m/s$$$
12.8 m/s