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?
