conventionsdragforcesnewtonian-mechanics
$$Drag = \frac{1}{2}C_d \rho Av^2$$
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.
In two dimensions, Newton's second law can be written in vector form as $$ \mathbf F_\mathrm{net} = m\mathbf a $$ In this case, the net force is $$ \mathbf F_\mathrm{net} = m\mathbf g - kv^2\frac{\mathbf v}{v} = m\mathbf g - kv\mathbf v $$ so the equation of motion is $$ m\mathbf a = m\mathbf g - kv\mathbf v $$ In components, if we choose the positive $y$ direction to be vertical, and using $v = \sqrt{v_x^2 + v_y^2}$ as you point out, we obtain $$ ma_x = -k\sqrt{v_x^2+v_y^2} v_x, \qquad ma_y = -mg-k\sqrt{v_x^2+v_y^2}v_y $$ as you can see, these differential equations are coupled; the $x$ equation involves $v_y$ and the $y$-equation involves $v_x$ unlike the case in which there is no drag. You should be able to numerically solve these simultaneous equations pretty easily on Mathematica.
In particular, you can solve these equations by specifying the initial position $\mathbf x(0) = (x(0), y(0)$ and the initial velocity $\mathbf v(0) = (v_x(0), v_y(0)) = (v(0)\cos\theta, v(0)\sin\theta)$ where $\theta$ is the initial angle at which the projectile is launched.
$C_d$ is a function of speed via the Reynolds number. See here and here.
For some numeric values for the $C_d$ vs. $\rm Re$ use the following from here of which you take the log values to interpolate.
The kinematic viscosity of Air is $\nu = 14.8\; \rm{cSt} = 14.8 \cdot 10^{−6}\; \rm{m^2/s}$. At a speed of $v=20\;\rm{m/s}$ a ball with diameter $d=5\,\rm{cm} = 0.05\;\rm{m}$ has Reynolds number of $\rm{Re} = \frac{v d}{\nu} = 67400 $.
When you look at the $C_d$ vs. $\rm Re$ graph you get $C_d = 0.47$. The area of the ball is $A=\pi \frac{d^2}{4} = 0.001964\;\rm{m^2}$ so the drag force is
$F_d = \frac{1}{2} \rho A C_d v^2 = \frac{1}{2} (1.2\;\rm{kg/m^3}) (0.001964\;\rm{m^2}) (0.47) (20\;\rm{m/s})^2 = 0.2215\;\rm{N}$
Best Answer
In short, the squared speed $v^2$ appears in the equation because when moving faster, you increase both
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.