Air drag on a vertically thrown object Air drag on a vertically thrown object is given directly propotional to its the square of its instantaneous velocity. But shouldn't it be linearly associated as air drag depends on amount of air displaced which is directly propotional to velocity?
 A: Suppose the stone has some cross sectional area $A$. If it's travelling at a velocity $v$ then in one second it sweeps out a volume $Av$. Therefore the mass of the air is displaces is:
$$ m_\text{air} = \rho Av $$
where $\rho$ is the density of the air.
For the next step we assume that the stone accelerates the air to match its own velocity so the change in the momentum of the air per second is:
$$ \Delta p_\text{air} = m_\text{air}v = \rho Av^2 $$
But the rate of change in the momentum is just the force, so we end up with:
$$ F = \rho Av^2 $$
This is an excessively simple calculation because in practice a moving object doesn't accelerate all the air it meets to match its own velocity. However it gives you a feel for where the $v^2$ term comes from.
A: You can get an intuitive sense for why there's a $v^2$ dependence by the following "toy model" analogy. Consider a large object, call it $M$, moving through absolutely "still air" comprised of a lot of "small marbles", each of the same mass $m$. The marbles are stationary (that's our "still air") and our object $M$ is moving through them with velocity $v$. So, each marble that $M$ collides with imparts a "retarding" momentum $mv$ to $M$ (actually, because $M\gg m$, each marble collision (assuming elastic collisions) really imparts $2mv$ to $M$, but that's a complication we can ignore for our purposes -- I only mention it to forestall comments).
So, if $M$ collides with $n$ such marbles in a second, then $M$'s momentum will be reduced by $n\times mv$. And that's linear in $v$. But now suppose you double $M$'s velocity. Not only will you double the effect of each collision ($2mv$ rather than $mv$), as above, but you'll double the number of marbles $M$ collides with each second ($2n$ rather than just $n$). So that's a factor of $2\times2=4$, and as I imagine you can see, it's $v^2$ (rather than just $v$) in general.
