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Homework Help: Terminal speed

  1. Feb 27, 2005 #1
    will a golf ball of mass 45g and diamter 4.3cm reach terminal speed when dropped from a height of 25m? the drag coefficient is 0.35 and the density of air is 1.2kg/m^3.

    using the formula [tex]v_t = sqrt(\frac{2*mg}{CpA})[/tex]

    [tex]v_t = sqrt(\frac{2*(45g)(9.8m/s^2)}{(.35)(1.2kg/m^3)(0.043m)})[/tex]

    so pluggeed that into my calculator, i found the terminal speed of 220.99m/s.
    this is where i got stuck, how would i know if it reaches terminal speed or not when dropping from a height of 25m?
  2. jcsd
  3. Feb 28, 2005 #2
    it will not reach the terminal speed nomatter how high you drop the ball......
    the velocity of the ball is......
    it will get closer and closer to the terminal speed when time passes, but it will never "reach" it
  4. Feb 28, 2005 #3
    i have a few questions, how did you get the formula v=v_t(1-e^(-kt))? and what does k and t stand for and how do i find it?
  5. Feb 28, 2005 #4
    solve the following DE
    [tex] -m\frac{dv}{dt}= -mg+ bv [/tex]
    b is the drag coefficient (i believe this is how ppl called it)
    k=b/m ...and t is time...
    Last edited: Feb 28, 2005
  6. Feb 28, 2005 #5
    sorry but a few more questions if you dont mind. how extactly would i find t? I dont think using one of the kenematics will help because i will be missing alot of info.
  7. Feb 28, 2005 #6
    you know calculus, do you?
    [tex] m\frac{dv}{dt}= mg- bv [/tex]
    [tex] \frac{dv}{dt}=g-\frac{b}{m}v[/tex]
    [tex] \frac{dv}{dt}=g-kv[/tex]
    [tex] dt = \frac{dv}{g-kv} [/tex]
    [tex]\int dt = \int \frac{dv}{g-kv} [/tex]
    [tex] t = \frac{-1}{k} ln(g-kv) + C [/tex]
    [tex] v(t) = \frac{g}{k}-\frac{e^{-k(t-C)}}{k} [/tex]
    [tex] v(t) = \frac{g}{k}-\frac{C'e^{-k(t)}}{k} [/tex]

    apply the initial condition [itex]v(0)=0[/itex] this implies [itex] C'=g [/itex]
    [tex] v(t) = \frac{g}{k}-\frac{ge^{-k(t)}}{k} [/tex]
    [tex] v(t) = \frac{g}{k}(1-e^{-k(t)}) [/tex]
    [tex] v(t) = \frac{gm}{b}(1-e^{-mt/b}) [/tex]

    whereas gm/b is the terminal velocity
  8. Feb 28, 2005 #7
    ah i see, thanks for the help!
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