Why is the drag force proportional to $v^2$ and defined with a factor of $1/2$? $$Drag = \frac{1}{2}C_d \rho Av^2$$
I understand that the strength of the drag depends on the density of the fluid the body passes through, the reference area of the body, the drag coefficient, and the velocity of the object.
I don't, however, understand the 1/2 and the $v^2$ in the equation.
 A: In short, the squared speed $v^2$ appears in the equation because when moving faster, you increase both


*

*how much momentum ($p=mv$) that is transferred to the air molecules (they must be moved away faster) and

*how much air that must be moved away (because you sweep through more air per second). 


Increasing the speed means increasing both of these factors that both make it tougher to fall. Thus, speed appears "twice", so to say.
The half $\frac 12$ that also appears in the equation, is - as others also point out - due to the drag coefficient $C_d$ being neatly written as $C_d=\frac{D}{Aq}$, where $q=\frac12\rho v^2$ is the dynamic pressure, an important aerodynamic property.
Sure, you could have included the half in the drag coefficient to simplify the drag formula. But you would simultaneously complicate the relationship $C_d=\frac{D}{Aq}$. You could also ask, why there is a half in $K=\frac12 m v^2$. Why isn't that half just included in the mass $m$? Well, because then many other relationships that include $m$ would become more complicated.
A: The drag force is doing negative work on the object that it is decelerating.  By the work/kinetic-energy theorem, the work done is equal to the change in kinetic energy that the object experiences.  Since kinetic energy is defined as $E_k = 1/2 mv^2$, you can expect the "1/2" and the $v^2$ terms to show up in the equation.
A: Adding just a bit to the previous answer, I believe that the drag coefficient definition is based on the dynamic pressure term in Bernoulli's equation, $\frac{1}{2}\rho v^2$. Thus dependence on velocity squared is expected, and is often observed. However, fluid flow is complicated, and $C_d$ determined empirically in many cases varies with flow velocity, and Reynolds number, depending on boundary layer transition from laminar to turbulent flow, form drag, wake, etc.
A: It's similar question if you would ask - Why kinetic energy is defined as $E_k = 1/2~m v^2$. The answer is below.
Elementary work is :
$$
\begin{align}
dW &= F \cdot dr 
\\&= F\frac {dr}{dt} dt
\\&=Fv~dt
\\&=m\frac {dv}{dt}v~dt
\\&=mv~dv 
\end{align}
$$
Now integrate both sides :
$$ \int dW = m \int v~dv  $$
Which gives :
$$W=E_k=1/2~mv^2$$
Thus the answer of why kinetic energy is proportional to $v^2$ and has half factor 1/2 in it, is that it is due to integration.
