*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:

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

We can 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 energy is equal to (dragging) force $[D]$ times distance:

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

Where we can rewrite the force $D$ as:

$$D = \frac{E}{ds}  \tag{5}$$

Therefore we can plug (3) into (5) to get:

$$D = \frac{E}{ds} = \frac{\frac{1}{2} \cdot \rho \cdot ds \cdot S \cdot V^2}{ds}  \tag{5} = \frac{1}{2} \cdot \rho \cdot S \cdot V^2$$

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 $m$. This is usually not the case, and to indicate to what degree this happens we add a correction factor $C_D$:

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

This also gives you a nice bridge to explain more about $C_D$ 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