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Terminal Velocity Equation 
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#1
Aug2804, 07:41 AM

P: 19

i'm looking for an equation to calculate terminal velocity. Does any one know it? and can u please post in here please.



#2
Aug2804, 08:52 AM

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PF Gold
P: 12,016

That would depend upon how you choose to model the air/fluid resistance.



#3
Aug2804, 10:09 AM

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PF Gold
P: 39,363

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+ kv^{2} and v satisfies the differential equation mdv/dt= g+ kv^{2}. That's a nonlinear differential equation but is separable and first order. We can integrate it by writing dv/(kv^{2}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. 


#4
Aug2804, 10:38 AM

P: 599

Terminal Velocity Equation
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 forceequations 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] 


#5
Jun2306, 07:53 PM

P: 1

Indeed, terminal velocity depends on mass. You forgot the FORCE due to gravity is mg, not g.



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