Acceleration of two falling objects with identical form and air drag but different masses I have a theoretical question that has been bugging me and my peers for weeks now - and we have yet to settle on a concrete answer.
Imagine two balloons, one is filled with air, one with concrete. They are both the same size (and hence have the same air resistance). On the moon, we are taught at school that they would drop at the same rate, however on earth, they obviously don't - with the concrete falling faster. 
Our reasoning was as follows: 
Air resistance is not a factor in this example, therefore the only other variable is the mass (density) of the two balloons. 
If you were to change the medium from earth's atmospheric air to water, the air ballon floats as it is less dense than water (and therefore displaces more water than its mass), and consequently has an upwards buoyancy force acting on it. The concrete sinks for the same reasons. 
Therefore in the medium of air, there is also, albeit somewhat reduced, bouyancy effect acting on the balloons - explaining why the concrete balloon falls faster. 
Could someone please tell me if my logic is correct, and if not explaing to us what other forces are playing their part.
Thank you so much for clearing this up! 
 A: Suppose the balloon with air (A) weighs $m_a$ and the balloon with concrete (B) weights $m_b$. The force accelerating the balloons downwards is $m_a g$ for A and $m_b g$ for B, where $g$ is the acceleration due to gravity. In the absence of air the acceleration is simply this force divided by the mass, so both balloons accelerate at the same rate of $g$. So far so good.
Now suppose the air resistance is F. We don't need to worry exactly what F is. The force on balloon A is $m_a g - F$, so its acceleration is
$$ a_a = \frac{m_a g - F}{m_a} = g - \frac{F}{m_a} $$
and likewise the acceleration of balloon B is
$$ a_b = \frac{m_b g - F}{m_b} = g - \frac{F}{m_b} $$
So the balloons don't accelerate at the same rate. In fact the difference in the accelerations is simply
$$ \Delta a_{ba} = a_b - a_a = \frac{F}{m_a} - \frac{F}{m_b} $$
Since $m_b \gg m_a$ the difference is positive, i.e. balloon B accelerates at a much greater rate than balloon A.
Response to comment
I think there are a couple of possible sources of confusion. Let me attempt to clarify these, hopefully without making things even more confused!
Firstly the air resistance affects both the acceleration and the terminal velocity. I've only discussed acceleration because terminal velocity can get complicated.
Secondly, and I think this is the main source of confusion, the gravitational force on an object depends on its mass while the air resistance doesn't. However the air resistance depends on the velocity of the object while the gravitational force doesn't. That means the two forces change in different ways when you change the mass and speed of the object.
Incidentally, Briguy37 is quite correct that buoyancy has some effect, but to keep life simple let's assume the object is dense enough for buoyancy to be safely ignored.
Anyhow, for a mass $m$ the gravitational force is $F = mg$ so it's proportional to mass and it doesn't change with speed. Since $a = F/m$ the acceleration due to gravity is the same for all masses.
By contrast the air resistance is (to a good approximation) $F = Av^n$, where $A$ is a constant that depends on the object's size and shape and $n$ varies from 1 at low speeds to 2 at high speeds. In your example the size and shape of the ballons is the same so $F$ just depends on the speed and for any given speed will be the same for the two ballons. The deceleration due to air resistance will depend on the object mass: $a = Av^n/m$, so air resistance slows a heavy object less than it slows a light object.
When you first release the balloons their speed is zero so the air resistance is zero and they would start accelerating at the same rate. On the moon there is no air resistance, so the acceleration is independant of speed and the two objects hit the ground at the same speed.
On the Earth the acceleration is initially the same for both balloons, but as soon as they start moving the air resistance builds up. At a given velocity the force due to air resistance is the same for both ballons because it only depends on the shape and speed. However because acceleration is force divided by mass the deceleration of the two balloons is different. The deceleration of the heavy balloon is much less than the deceleration of the light balloon, and that's why it hits the floor first.
From the way you phrased your question I'm guessing that and answer based on calculus won't be that helpful, but for the record the way we calculate the trajectory of the falling balloons would be to write:
$$ \frac{dv}{dt} = g - \frac{Av^n}{m} $$
This equation turns out to be hard to solve, and we'd normlly solve it numerically. However you can see immediately that the mass of the object appears in the equation, so the change of velocity with time is affected by the mass.
A: You don't need buoyancy to explain what you are seeing...
Your formula here is $ma = mg - R$, so $a = g - R/m$, where $R$ is the air resistance. If it is the same for both objects, you can see that, the more massive the object is, the least it will affect its acceleration.
A: I believe the explanations based on air resistance alone to explain this are incorrect.  My reasoning is that if you threw both balloons up in the air, at the top of their arcs there would be no air resistance because air resistance is a function of relative air speed and would thus be zero at that point, yet the air-balloon accelerates much slower than the concrete balloon even at that point.  On top of this, even when air resistance is working with gravity (on the way up), the balloon filled with air still accelerates downward much slower than the balloon filled with concrete.*
Thus, I believe you are correct to assume that buoyancy is the reason for this.  To show this better, let's do an example and say there is a cubic foot of air in the balloon at sea level, which equates to 0.0807 lbs or 37 grams (let's not take into account the added pressure on the air inside the balloon which would make it slightly more dense).  Next, let's say that the deflated latex balloon weighs 3 grams and takes up a negligible amount of space when considered with the space the blown-up balloon takes.  
From the above setup, we have a total mass of 40 grams in the blown-up air-balloon floating around where there is normally a stable mass of 37 grams of air.  At this point, the force of gravity on the 37 grams essentially cancels out since air of that weight in the same space floats, and thus you are left with only the force of gravity on the 3 grams that affects the acceleration of the balloon.  Thus, your base acceleration in this case would be:
acceleration = force / mass = 3grams * g / 40grams = .075 * g = .075 * -9.8 m/s = -.735 m/s/s

This seems pretty close to the acceleration I have observed from my mass amounts of experience with bouncing balloons on my hands to keep them aloft.  Of course, once your balloon starts moving more in relation to air, then the air resistance becomes the limiting factor.
Also, let's consider the concrete balloon.  If you assume a cubic foot concrete balloon is 140 lbs, or 63,502 grams without the 3 gram balloon weight, acceleration using the same formula would be:
acceleration = 63,502 grams * g / 63,505 grams = .999953 * g

Which is close enough for us to call g, or -9.8m/s/s, so you can see why it accelerates so much faster.
Hope this helps!
Notes:
* With quick speeds, air resistance on the way up becomes the driving factor on the air-balloon, and can thus accelerate it much faster than gravity can slow the concrete balloon.  Thus, this statement assumes that the air balloon is moving slowly enough in the upward direction where gravity is the driving force on the acceleration of the balloon.  
A: Concrete balloon is massive (more inertia), so there's require more air resistance (air drag, buoyant force whatever) to reduce, stop or reverse its motion than that of for an air balloon. As the air resistance is same for both, concrete ballon will fall faster. So, simple.. no Rocket Science.
