# About air resistance in horizontal direction

1. Jul 11, 2011

### amiras

Hello, I was playing with formulas on air resistance and derived formula for velocity in horizontal direction after object was thrown. It looks reasonable but I started have confusion about it, i'd like for you great experts check on its derivation, so here it goes every v is v in x direction.

So I assumed that then object is thrown the only force that acts on it in x direction is air resistance opposite to its velocity, f = Dv^2, where D some constant.

so it should be: f = ma (since the force is in the same direction as acceleration (opposite to object velocity)), but I skipped this part and wrote:

-f = m dv/dx * dx/dt = mv dv/dx

-Dv^2 = mv dv/dx

-D/m dx = dv/v

and after integrating x from 0 to x, and from v0 to v, I've got: v = v0 * e^(-Dx/m)

It looks reasonably to me since velocity always decrease from its maximum value (v0), how its suppose to be.

Now after a while I got confused about this part: -f = mv dv/dx
if f = ma, according to me f $\neq$ mv dv/dx

Is it right because change in velocity over distance is in opposite direction as the force (and acceleration)? But velocity is also in different direction so why it is not like f = m(-v)(-dv/dx)

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