homework and exercises - Splitting up a force into horizontal and vertical components?

My Bedford and Fowler textbook (4th edition) has a chapter on numerical solutions. I'm having trouble understanding how the textbook splits up the components of force in the $x$ and $y$ directions to model the drag on a projectile:

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The aerodynamic frag force on the projectile is of magnitude $C|{v^2}|$, where $C$ is a constant.

We must determine the $x$ and $y$ components of the total force on the projectile. Let $D$ be the drag force. Because $v/|v|$ is a unit vector in the direction of $v$, we can write the drag force as:

$$D = -C{|v|^2} \frac{v}{|v|} = C|v|v.$$

The external forces on the projectile are its weight and the drag so we have:

$$\sum F = -mgj - C|v|v,$$

and the total components of the force are:

$$\sum F_x = C\sqrt{v_x^2 + v_y^2}\quad v_x$$

$$\sum F_y = -mg- C\sqrt{v_x^2 + v_y^2}\quad v_y.$$

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I don't understand how the individual components of the drag force $D = C|v|^2$ (acting in opposition to the projectile's movement) becomes something like this:

$$-C\sqrt{v_x^2 + v_y^2}* v_x~ ?$$

I understand that the absolute value of something can be represented as:

$$|a|=\sqrt{a^2}.$$

And the magnitude of velocity from its $x$ and $y$ components can be calculated as

$$V=\sqrt{v_x^2+v_y^2}.$$

But I don't get how the non-absolute $v$ term magically becomes $v_x$ or $v_y$ while the absolute term doesn't become $|v_x|$ or $|v_y|$ too.

What about equations without absolute values in them like the drag force equation? $F_D = \frac{1}{2}pv^2AC_D$, where $p,A, C_D$ are constants. Would the horizontal and vertical components of forces then be:

$$F_x = -\frac{1}{2}pv_x^2AC_D$$ and $$F_y = -mg -\frac{1}{2}pv_y^2AC_D~ ?$$

Answer

I find this type of question is always easier if you draw a diagram. The drag force $F$ acts in the opposite direction to the velocity so it looks like:

and the components of the drag force are:

$$\begin{align} F_x &= F \cos\theta \\ F_y &= F \sin\theta \end{align}$$

$\cos\theta$ is $v_x/v$ and $v = \sqrt{v_x^2 + v_y^2}$ so we get:

$$F_x = F \frac{v_x}{\sqrt{v_x^2 + v_y^2}}$$

Since $F = -Cv^2$ we get:

$$F_x = -Cv^2 \frac{v_x}{\sqrt{v_x^2 + v_y^2}}$$

and because $v = \sqrt{v_x^2 + v_y^2}$ this becomes:

$$\begin{align} F_x &= -C \left(v_x^2 + v_y^2\right) \frac{v_x}{\sqrt{v_x^2 + v_y^2}} \\ &= -C v_x \sqrt{v_x^2 + v_y^2} \end{align}$$

And likewise for $F_y$.