Two balls A and B of the same size are dropped from the same point. If the mass of A is greater of the two, and if air resistance is the same on both, which ball will reach the ground first/simultaneously?
I thought that since the acceleration acting on them is same, both will reach the ground simultaneously. But the answer in the book says that ball A (the one having greater mass) will reach first. I mean, isn't this what people thought before (I guess it's true) Galileo performed his experiment, until they were proven wrong?!
 A: 
... since the acceleration acting on them is same

Acceleration cannot be "acting" on anything. Forces act on objects. There are two forces: gravitational $F_g = mg$ and air resistance $F_{air} = const$.
The total force on an object of mass $m$ will be $F = mg - F_{air}$.
Thus the acceleration will be $a = g - \frac{F_{air}}{m}$
A: The heavier one will reach the ground first. Note that saying that the air resistance is the same for both is a bit vague, and it is by no means the same thing as neglecting air resistance. The drag force depends on the shape of an object (surface area) and therefore it is the same for two identical balls with different mass. But the acceleration experienced by the body is inversely proportional to its mass. Thus you have
$$\vec{F}=\vec{F}_g+\vec{F}_{drag}$$
Now the gravitational force has the very special property of being proportional to the mass of the body itself ($\vec{F}_g=m\vec{g}$), and so the equation of motion is:
$$\ddot{y}=g-\frac{F_{drag}}{m}$$
As you can see the second term on the right hand side becomes less and less important as the mass of the body is increased, resulting in a larger acceleration. If one neglects air resistance, on the other hand, such term vanishes no matter the value of $m$, and so the bodies fall at the same rate as in the experiment performed by Galileo. 
