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Need help with an Integration

  1. Sep 20, 2004 #1
    What is the technique for integrating velocity with respect to dy?

    In other words:
    integral(dv^2/dt) dy [ from 0 to h ]

    if anyone can help, that'd be super.. mkay
  2. jcsd
  3. Sep 20, 2004 #2


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    Without knowing, the relationship of velocity to y, no!

    You don't even say, though I would guess it, the the velocity is the velocity of something in the y direction- that is, that v= dy/dt. d(v2)/dt= 2 v dv/dt.
    We still can't integrate that with respect to y until we have some idea of the relationship between y and t.
  4. Sep 20, 2004 #3
    at time t = 0, y = 0
    at some time t, y will be a maximum.

    The problem reads: Find the speed of a bullet fired straight up when the resistive force is given by F(v) = -cv^2
    so you have

    1/2 mdv^2/dy = -mg - cv^2
    mdv^2/dy = 2 (-mg -cv^2)

    which is a seperable differential equation

    mdv^2 = 2 (-mg - cv^2) dy

    dv^2 = (-2g - 2cv^2/m) dy = -2 (g + cv^2/m) dy

    so... this is where I am a little stuck
    can you suggets where to proceed?
  5. Sep 21, 2004 #4


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    The whole point of a "separable equation" is to separate the variables!

    Once you have md(v2)= 2 (-mg - cv2) dy

    Rewrite it as [itex] m\frac{d(v^2)}{cv^2+ mg}= -2ydy[/itex] and integrate.

    (I started to write d(v2) as 2v dv but then I recognized that you have only v2 and not v itself in the integral. If you let y= v2, the integral on the left is exactly the same as the integral of [itex]m\frac{dy}{cy+mg}[/itex].)
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