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Air resistance formula

  1. Sep 13, 2005 #1
    "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.
     
  2. jcsd
  3. Sep 13, 2005 #2
    Does anyone know how to do this? I really need help on this one and it's due tomorrow.
     
  4. Sep 13, 2005 #3
    ok, you have

    [tex]
    m\dot{v}=mg-bv^2
    [/tex]

    this will be alot easier to do if you make the substitution

    [tex]
    v_t =\sqrt{\frac{mg}{b}}
    [/tex]

    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

    [tex]

    \dot{v}= g(1-\frac{v^2}{v_t^2})
    [/tex]

    then from here you can use seperation of varbs to get

    [tex]

    \frac{dv}{1-\frac{v^2}{v_t^2}}=gdt
    [/tex]

    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).
     
  5. Sep 14, 2005 #4
    Thanks man. That helps a lot. Now if only I'd studied hyperbolic functions...
     
  6. Sep 14, 2005 #5
    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.
     
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