Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Science Advisor

    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


    User Avatar
    Science Advisor

    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].)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook