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. 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 $C_D$?
