Advanced Quadratic drag - How to solve?

So I have gotten this far...

#a=((F_a-(c_d*rho_h*A*v^2)/2)-m_t*g_r)#

This is based upon #F_a-(F_d+F_g)/m =a#

But #a# is #(dv)/(dt)#

So this becomes a differential equation because the faster you go the harder it is to go faster.

How can one solve this?

Please note I'm currently a Calc AB student.

1 Answer
Jul 23, 2016

Answer:

see below

Explanation:

your work looks a little bit messed up. EG It is dimensionally incorrect for starters.

I can see the sense of it, the interaction between Rayleigh drag and gravity. Is there also an additional force that is purporting to act on the mass? i can't figure that out on a quick look.

the DE you get shouldn't be too much trouble, in the sense it should be separable. the problem you will have is the same as with all damping/ friction like effect in that it opposes motion and so you becomes obsessed with the actual sign of v

if you linked the actual prob, that would be helpful

FWIW if you have a situation where a body was free falling in drag, you could say in the vertical x direction, with downward as positive co-ordinate, that

and #F = mg - 1/2 \ rho v^2 \ C_D \ A = m (dv)/dt#

that is separable though the solution might not be so sweet

post the question or some more specifics. hope that helps in some way