How can air resistance be incorporated into the falling particle equation?

[pi-r8]

"A particle is relased from rest (y = 0)and falls under the influence of gravity and air resistance. Find the relationship between v and the distance of falling y when the air resistance is equal to a) av and b) Bv^2."

I already solved part a by integrating dv/dt = -g - (a/m)(dy/dt) with respect to t to give me v = -gt - (ay/m), and then taking force = m(dv/dt) = -mg-av and rearranging this to give me dv/(g + av/m) = -dt, which I integrated to get (m/a)ln(av/m + g) = -t + c. Solving for c and doing some algebra gives me y = -(m/a)(v -(mg/a)ln(av/mg + 1)) as my final solution for y.

Part b, though, I have no idea how to do. If I try and solve it the same way, I end up with the integral of dv/(g + av^2/m) with respect to time, which I have no idea how to solve. I also have the integral -g-(av^2)/m with respect to time, which I don't know how to solve either. If anyone can help me solve these integrals, or point out an easier way to solve this problem, I'd be most grateful.

 

[pi-r8]

Does anyone know how to do this? I really need help on this one and it's due tomorrow.

 

[matpo39]

ok, you have

$$m\dot{v}=mg-bv^2$$

this will be a lot easier to do if you make the substitution

$$v_t =\sqrt{\frac{mg}{b}}$$

where v_t is the terminal velocity which is when the right hand side of the eq. balances out. subing in v_t we get

$$<br /> \dot{v}= g(1-\frac{v^2}{v_t^2})$$

then from here you can use separation of varbs to get

$$<br /> \frac{dv}{1-\frac{v^2}{v_t^2}}=gdt$$

and simply integrat over v and t.
it may be very helpful to look at the hyperbolic functions most notably that of the arctanh(x).

 

[pi-r8]

Thanks man. That helps a lot. Now if only I'd studied hyperbolic functions...

 

[shaner-baner]

You don't need hyperbolic functions, although that is an alegant way to do it. Another option is to factor the denominator into a sum and difference and then apply partial fraction decomposition. The resulting fractions only have linear denominators and can easily be integrated.