Newton's Mechanics

Let the masses of A and B be $M$ and $m$ respectively (with $M > m$ ). The net force of each object as they're thrown upward is equated by:

$Ma_{A} = Mg + F_{air} \Rightarrow a_{A} = g + \frac{F_{air}}{M};$

$ma_{B} = mg + F_{air} \Rightarrow a_{B} = g + \frac{F_{air}}{m};$

which upon observation $a_{A} < a_{B}$ . Now the heights of each object can be modeled via uniform acceleration kinematics:

$v_{final}^{2} = v_{initial}^{2} + 2ay$

with $v_{initial} = v_0, v_{final} = 0$ for both cases. Solving for $y$ in either scenario gives:

$y_{A} = \frac{v_0^{2}}{2a_{A}}$ and $y_{B} = \frac{v_0^{2}}{2a_{B}}$

and since $a_{A} < a_{B}$ , this results in $\boxed{y_{A} > y_{B}}.$