1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Terminal Velocity Equation

  1. Aug 28, 2004 #1
    i'm looking for an equation to calculate terminal velocity. Does any one know it? and can u please post in here please. :smile:
  2. jcsd
  3. Aug 28, 2004 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    That would depend upon how you choose to model the air/fluid resistance.
  4. Aug 28, 2004 #3


    User Avatar
    Science Advisor

    One common model is that the resistance force is proportional to the speed.
    Under that model, an object falling, under gravity has acceleration -g+ kv (k is the proportionality constant, v the speed. Since that is a function of v, it give the linear differential equation mdv/dt= -mg+ kv. The general solution to that is v(t)= Ce-kt/m-mg/k. For very large t, that exponential (with negative exponent) goes to 0 and the "terminal velocity" is -mg/k.

    Another common model is to set the resistance force proportional to the square of the speed. That means the net force is -g+ kv2 and v satisfies the differential equation mdv/dt= -g+ kv2. That's a non-linear differential equation but is separable and first order. We can integrate it by writing
    dv/(kv2-g)/m= (-1/2√(g))(1/(√(k)v+√(g))dv/m+(1/2√(g))(√(k)v-&radic(g))dv/m= dt.
    Integrating both sides, we get (1/2√(kg))ln((√(k)v-√(g))/(√(k)v+√(g))= mt+ C. For large t, the denominator on the left must go to 0: the terminal velocity is -√(g/k) which, you will notice, is independent of m. This model is typically used for very light objects falling through air or objects falling through water.
  5. Aug 28, 2004 #4
    A falling object on earth is subjected to a downward force [itex]F_g=mg[/itex], while it's air resitance constitutes an upward force often modelled by

    [tex]F_w=\frac{1}{2}C_D \rho A_F v^2[/tex].

    With [itex]C_D[/itex] a constant (drag coefficient) that models how aerodynamic the object is, for most object of the order 1 (for a raindrop for example ~0,5), [itex]\rho[/itex] the air density, [itex]A_F[/itex] the frontal area of the object (perpendicular to the direction of motion), and [itex]v[/itex] the velocity of the object.

    At terminal velocity the force on the object is zero (otherwise the object would accellerate!) so you can equate both force-equations yielding:

    [tex]mg=\frac{1}{2}C_D \rho A_F v^2[/tex]
    [tex]v=\sqrt{\frac{2mg}{C_D \rho A_F}}[/tex]
  6. Jun 23, 2006 #5
    Error in post 3

    Indeed, terminal velocity depends on mass. You forgot the FORCE due to gravity is mg, not g.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook