Which ball touches the ground first? This is a very well known problem, but I can't find an answer in the specific case I'm looking for.
Let's consider two balls :


*

*Ball 1 weighs 10 kg

*Ball 2 weighs 1 kg

*Balls have identical volumes (so Ball 1 is much more dense)

*Balls have identical shapes (perfect spheres)


Let's drop them from a rather important height, on earth, WITH air. (That's the important thing, because all the proofs that I browse take place in a vaccum).
I am arguing with a colleague. He thinks that ball 1 will fall faster in air, and that the two balls will fall at the same speed in a vacuum. I think that the identical shapes and volumes make air friction identical too and that the vaccum has no importance here. Could someone tell who's right and provide a small proof?
 A: I am sorry to say, but your colleague is right.
Of course, air friction acts in the same way. However, the friction is, in good approximation, proportional to the square of the velocity, $F=kv^2$. At terminal velocity, this force balances gravity,
$$ m g = k v^2 $$
And thus
$$ v=\sqrt{\frac{mg}{k}}$$
So, the terminal velocity of a ball 10 times as heavy, will be approximately three times higher. In vacuum $k=0$ and there is no terminal velocity (and no friction), thus $ma=mg$ instead of $ma=mg-F$.
A: Ball 1 will drop faster in air, but both balls will drop at the same speed in vacuum.
In vacuum, there is only the gravitational force on each ball.  That force is proportional to mass.  The acceleration of a object due to a force is inversely proportional to its mass, so the mass cancels out.  Each ball will accelerate the same, which is the accelleration due to gravity for the local conditions (about 9.8 m/s2 on the surface of the earth).
However, in air there is the additional upwards force due to friction with the air.  That force is a function of the speed and shape of the falling object.  If both balls were falling at the same speed, both would have the same upwards force on them due to air resistance.  This force is not proportional to the mass of the object, so causes a higher deceleration on the object with less mass.
For example, the 10 kg ball is pulled downwards due to gravity with 98 N force, whereas the 1 kg ball is only pulled downwards with 9.8 N.  Let's say they are falling at the same speed thru air and that each is experiencing 3 N upwards force due to the air.  Ball 1 is now being pulled down by a total of 95 N, and ball 2 by 6.8 N.  That means ball 1 experiences 95 N / 10 kg = 9.5 m/s2 downward acceleration, and ball 2 experiences 6.8 N / 1 kg = 6.8 m/s2 downward acceleration.  This means ball 1 will continue on falling faster than ball 2.
A: I am not satisfied with the way @Bernhard answered, since it just shows the maximum velocity, thus only answering partially the question.
The air resistance can be written as :
$$
R = \frac{1}{2}\,C_x\, \rho\, S\, v^2
$$
Note : The mass of the object is not in this equation. This is very important.
Applying Newton's law to one of the object gives at any moment of the fall:
$$
a = g - \frac{1}{2m}\,C_x\, \rho\, S\, v^2
$$
As you can see the acceleration is function of the mass of the object $m$. A heavier object will accelerate more than a lighter one, therefore, will go faster during the whole fall. Both objects will at one point reach the maximum velocity that is explained well in @Bernhard answer.
So at any point of the fall, your heavier object will be faster than the lighter one.
A: Other answers & comments cover the difference in acceleration due to drag, which will be the largest effect, but don't forget that if you are in an atmosphere there will also be buoyancy to consider. 
The buoyancy provides an additional upward force on the balls that is equal to the weight of the displaced air. As it is the same force on each ball, the acceleration resulting from this force will differ based on the mass of the ball.
This is most easily illustrated by considering one as a lead ball and one as a helium balloon - obviously the helium balloon doesn't fall, because it is lighter than the air it displaced. The upward buoyancy force is greater than the downward gravitational force.
In a heavier fluid, like water, this effect is even more pronounced.
A: Since air creates a force that is approximately proportional to the square of the velocity, the acceleration for each sphere is $a_r = kv^2/m (where \text{ }  k = \frac{1}{2} C_x\rho\ S) $ The net acceleration on each sphere is $ a_n = g - a_r$. As the velocity increases, the $a_r $ increases until the net acceleration $a_n $ becomes zero $(a_r = g)$, and thus each sphere reaches its terminal velocity. $$Given:  m_1 = 10kgr, \text{ } \text{ } m_2 =1kgr, \text{ } k = 0.01, \text{ }g = 9.8m/s$$ $$  For \text{ } m_2, ( v_2 = m_2g/k)^{1/2} = (1x9.8/.01)^{1/2} = 31.3 m/s$$ $$For \text{ } m_1, (v_1 = m_1g/k)^{1/2} = (10x9.8/.01)^{1/2} = 98.99m/s$$
After using an iterative method, I determined that the 1kgr mass $(m_2)$ reaches the terminal velocity in about 10 seconds and the 10kgr mass $m_1$ in about 33 seconds.  Although the spheres reach their terminal velocity at different times, the larger mass reaches a higher velocity because the lighter mass reaches its terminal velocity sooner and does not increase after that.  The heavier mass, takes longer to reach its terminal velocity, and thus it becomes larger.  So, the heavier mass will reach the ground sooner.  
A: This problem can be easily solved by the formula “F=ma”.
You must be familiar with the reason why it would fall at the same rate in vacuum.
But if we talk about the free fall in atmosphere, as you said there will be friction off course, and as the objects have the same shape, it’ll be same. 
As the force of friction is the same on the two bodies, the one with the larger mass will have a smaller (negative) acceleration , and the one with the smaller mass have a bigger (negative) acceleration. So the ball with smaller mass will be slowed down by a great extent ( than the ball with larger mass). 
ALWAYS remember, F=ma. Force depends ONLY on the mass and NOT on the density!
PS - I don’t know why are others making the problem so complicated with those formulas! 
