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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


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


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    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.
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