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By forming the ratio of the drag to weight, you find that it is exactly proportional to the ratio of the velocity to the terminal velocity. This gives the result that linear drag is negligible for velocities much less than the terminal velocity. For example, a velocity which is 1% of the terminal velocity implies a drag which is 1% of the weight.

In the case of quadratic air resistance, a similar manipulation allows you to see that the ratio of drag to weight goes with the square of the ratio of the velocity to the terminal velocity. For a velocity which is 10% of the terminal velocity this implies that the drag is 1% of the weight. So for velocities which are much less than the terminal velocity, the drag is even more negligible in the case of quadratic air resistance.

This seems so counterintuitive and almost paradoxical to me. Taking air resistance as a function of velocity, it is clear that for velocities greater than 1 m/s the quadratic air resistance will provide a stronger force. Despite supplying greater force, it seems that for velocities less than the terminal velocity, quadratic air resistance gives you "less bang for your buck", so to speak.

If anyone has any useful insight about this situation I would love to hear it.

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Taking air resistance as a function of velocity, it is clear that for velocities greater than $1~\rm m/s$ the quadratic air resistance will provide a stronger force.

This seems to come from a misconception that $v^2 > v$ if $v > 1~\rm m/s$. In reality, neither of $4~\rm m^2/s^2$ or $2~\rm m/s$ are larger than the other- they have different units and so can't be compared. The quadratic term looks larger, but in another system of units it might be the other way around.

The drag force as a function of velocity can be given by $F(v)=-av-bv^2$, where $a$ and $b$ are two different constants that depend on the object being considered. Depending on the actual values of $a$ and $b$, the relative strengths of the two terms can vary.

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