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Vertical motion with retarding force

  1. Jan 26, 2008 #1
    Hello everyone, I've used these forums quite a bit and found it very helpful thanks for all you kind souls helping us through the sciences, but today is my first post :P

    I am deriving the velocity of a projectile fired vertically through a retarding force and I continue getting every step the same as the book until the last equation, it may be due to it being nearly 2 in the morning but I wanted to see others thoughts on it to help me learn the reasoning better. (And oh yeah this retarding force is only linear to the velocity)

    so if we have:

    -mg - km[tex]\dot{y}[/tex] = m[tex]\ddot{y}[/tex]

    we can easily work to

    -g - k[tex]\dot{y}[/tex] = [tex]\ddot{y}[/tex]

    [tex]\frac{dv}{dt}[/tex] = -g - kv

    dv = dt(-g - kv)

    [tex]\int\frac{dv}{g + kv}[/tex] = -[tex]\int{dt}[/tex]

    [tex]\frac{1}{k}[/tex] ln(g + kv) = -t + c

    ln(g + kv) = -kt + c

    g + kv = [tex]e^{-kt + c}[/tex]

    This is where the book and I agree to

    but then it arrives to

    v = [tex]\frac{dy}{dt}[/tex] = -[tex]\frac{g}{k}[/tex] + [tex]\frac{kv_0 + g}{k} e^{-kt}[/tex]

    Can someone explain to me how it arrives there? Thank you so much!
  2. jcsd
  3. Jan 26, 2008 #2

    Just notice, you can find c when t = 0, v = v0:

    ln(g + kv) = -kt + c =>
    ln(g + kv0) = -k.0 + c <=>
    c = ln(g + kv0)

    g + kv = [tex]e^{-kt + c}[/tex] <=>
    g + kv = [tex]e^{-kt + ln(g + kv0)}[/tex] <=>
    g + kv = [tex]e^{-kt}[/tex].(g + kv0)
  4. Jan 26, 2008 #3
    ah ok perfect thanks a bunch I kept going around in circles with it last night.
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