We apply the work-energy theorem, which states that the total work done on the object, $W_{total}$, is equal to the change in its kinetic energy, $\Delta K$. The total work is the sum of the work done by gravity, $W_g$, and the work done by air resistance, $W_{air}$.

The change in kinetic energy is:

$$\Delta K = K_f - K_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2$$

The work done by the conservative gravitational force depends only on the change in vertical position. Taking the ground as the reference level ($y=0$):

$$W_g = -\Delta U_g = -(U_{g,f} - U_{g,i}) = -(0 - mgh) = mgh$$

According to the work-energy theorem:

$$W_{total} = W_g + W_{air} = \Delta K$$

Substituting the expressions for $W_g$ and $\Delta K$:

$$mgh + W_{air} = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_0^2$$

Solving for the work done by air resistance, $W_{air}$:

$$W_{air} = \frac{1}{2}m(v_f^2 - v_0^2) - mgh$$