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**TL;DR Summary:**I have a water drop falling with a constant velocity ##v##, since ##mg=\frac 12 C \rho S v^2##. Wind is blowing with a velocity ##u## only in a horizontal direction. What will be its force acting on the drop?

I would use this equation ##F= \frac 12 C\rho S u^2##. When I want to calculate the net velocity of the drop, will it be the sum of ##\vec v## and ##\vec u##? Bacause we assume that the drop is so light that its horizontal velocity will be equal to the wind's one.

So the drop's velocity should be then constant, but when threre is a force ##F## acting on it, how is it possible? Doesn't act on the drop also a force ##-F##, the force caused by a normal air resistance?