*This explanation assumes that the high school students know the concepts of kinetic energy and work*

If we look at the kinetic energy of the air in front of the object, we note that:

$$ E = \frac{1}{2} \cdot m \cdot V^2 \tag{1} $$

If we assume that only the air within distance $ds$ times a frontal surface $S$ gets affected, we can write $m$ as:

$$ m = \rho \cdot ds \cdot S  \tag{2}$$

If we plug $(2)$ into $(1)$ to get:

$$E = \frac{1}{2} \cdot \rho \cdot ds \cdot S \cdot V^2 \tag{3}$$

We also know that work is equal to (dragging) force $[D]$ times distance:

$$ W = D \cdot ds \tag{4} $$

Due to the energy balance, the energy lost by the air $E$ is equal to the work $W$ done on the object :
$$W = E \tag{5}$$
We then subsitute $(4)$ for the left hand side and $(3)$ for the right hand side:
$$ D \cdot ds  =  \frac{1}{2} \cdot \rho \cdot ds \cdot S \cdot V^2  \tag{6} $$
And we can divide both left and right by $ds$ to obtain:
$$ D  =  \frac{1}{2} \cdot \rho  \cdot S \cdot V^2  \tag{7} $$

However, this assumes that all the energy contained within our air packet $m$ is completely transferred to the object. And the influence of the object is indeed limited to our air packet $m$ (with size $ds \times S$) . This is usually not the case, and to indicate to what degree this happens we add a correction factor $C_D$ to $(7)$:

$$ D = C_D \cdot \frac{1}{2} \cdot \rho \cdot S \cdot V^2 $$

This also gives you a nice bridge to explain more about the meaning $C_D$ and how it varies for different shapes.
In what situation will you have a $C_D$ of 1? What can you do to reduce $C_D$? What does it mean when $C_D$ is larger than 1?

[![enter image description here][1]][1]

[Image from very useful Wikipedia page of Drag Coefficient][2]


  [1]: https://i.sstatic.net/xtHW3.png
  [2]: https://en.wikipedia.org/wiki/Drag_coefficient