Not sure this is a good example but imagine we have two feathers, both with exactly the same drag coefficient, they have the exact same shape and everything, but the only difference is that one of them is somehow much more heavier (let's say it's much more dense), would their terminal velocities be different?
[Physics] If two objects have the same drag coefficient, but one is much heavier, would it fall faster
dragmassnewtonian-gravitynewtonian-mechanicsprojectile
Related Solutions
Using your definition of "falling," heavier objects do fall faster, and here's one way to justify it: consider the situation in the frame of reference of the center of mass of the two-body system (CM of the Earth and whatever you're dropping on it, for example). Each object exerts a force on the other of
$$F = \frac{G m_1 m_2}{r^2}$$
where $r = x_2 - x_1$ (assuming $x_2 > x_1$) is the separation distance. So for object 1, you have
$$\frac{G m_1 m_2}{r^2} = m_1\ddot{x}_1$$
and for object 2,
$$\frac{G m_1 m_2}{r^2} = -m_2\ddot{x}_2$$
Since object 2 is to the right, it gets pulled to the left, in the negative direction. Canceling common factors and adding these up, you get
$$\frac{G(m_1 + m_2)}{r^2} = -\ddot{r}$$
So it's clear that when the total mass is larger, the magnitude of the acceleration is larger, meaning that it will take less time for the objects to come together. If you want to see this mathematically, multiply both sides of the equation by $\dot{r}\mathrm{d}t$ to get
$$\frac{G(m_1 + m_2)}{r^2}\mathrm{d}r = -\dot{r}\mathrm{d}\dot{r}$$
and integrate,
$$G(m_1 + m_2)\left(\frac{1}{r} - \frac{1}{r_i}\right) = \frac{\dot{r}^2 - \dot{r}_i^2}{2}$$
Assuming $\dot{r}_i = 0$ (the objects start from relative rest), you can rearrange this to
$$\sqrt{2G(m_1 + m_2)}\ \mathrm{d}t = -\sqrt{\frac{r_i r}{r_i - r}}\mathrm{d}r$$
where I've chosen the negative square root because $\dot{r} < 0$, and integrate it again to find
$$t = \frac{1}{\sqrt{2G(m_1 + m_2)}}\biggl(\sqrt{r_i r_f(r_i - r_f)} + r_i^{3/2}\cos^{-1}\sqrt{\frac{r_f}{r_i}}\biggr)$$
where $r_f$ is the final center-to-center separation distance. Notice that $t$ is inversely proportional to the total mass, so larger mass translates into a lower collision time.
In the case of something like the Earth and a bowling ball, one of the masses is much larger, $m_1 \gg m_2$. So you can approximate the mass dependence of $t$ using a Taylor series,
$$\frac{1}{\sqrt{2G(m_1 + m_2)}} = \frac{1}{\sqrt{2Gm_1}}\biggl(1 - \frac{1}{2}\frac{m_2}{m_1} + \cdots\biggr)$$
The leading term is completely independent of $m_2$ (mass of the bowling ball or whatever), and this is why we can say, to a leading order approximation, that all objects fall at the same rate on the Earth's surface. For typical objects that might be dropped, the first correction term has a magnitude of a few kilograms divided by the mass of the Earth, which works out to $10^{-24}$. So the inaccuracy introduced by ignoring the motion of the Earth is roughly one part in a trillion trillion, far beyond the sensitivity of any measuring device that exists (or can even be imagined) today.
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.
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Best Answer
Imagine the lighter object falling with a terminal speed.
Whilst this is happening the drag force upwards is equal to the weight of the lighter object downwards.
Now consider the heavier object travelling at the same speed as the lighter object's terminal speed.
The drag force upwards is the same whilst the weight of the heavier object is greater than the weight of the lighter object and hence the drag force.
So the heavier object must accelerate downwards eventually reaching a higher terminal speed than the lighter object.