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Need help with ballistic equations

  1. Sep 23, 2011 #1
    [URL]http://latex.codecogs.com/gif.latex?(d^2Y)/(dTdY)=%20-c(dY/dT)-9.8(dT/dY)[/URL]

    basically how do you integrate dy/dt with respect to y, I know dy/dt integrated with respect to t is simply Y, but the other I have no idea.

    background: C is a constant that is a function of air pressure and is from the drag equation.
     
    Last edited by a moderator: Apr 26, 2017
  2. jcsd
  3. Sep 23, 2011 #2
    the equation you posted doesn't make sense to me. but to answer your second question
    "how do you integrate dy/dt with respect to y"
    you should know that dy/dt = 1/(dt/dy) so you can integrate with respect to y if you can write t=t(y).
     
  4. Sep 23, 2011 #3
    actually I re wrote the thing in an easier form but it gets messy if you just use separation of variables

    dv/dt=-fV^2-9.8

    I can't remember how to do this using ODE, any help?
     
  5. Sep 23, 2011 #4

    BruceW

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    I'm having trouble understanding your first problem.
    [tex]\frac{d^2Y}{dTdY}=-c \frac{dY}{dT} -9.8 \frac{dT}{dY} [/tex]
    The bit on the left-hand side could be rearranged:
    [tex]\frac{d}{dT} ( \frac{dY}{dY} )[/tex]
    which is equal to zero, right? So then you'd have:
    [tex] (\frac{dY}{dT})^2 = \frac{-9.8}{c} [/tex]
    Does this look right? Are you trying to model an actual physical process, or is it just a maths problem?

    And your second equation:
    [tex]\frac{dv}{dt}=-f V^2 -9.8[/tex]
    Is f a constant or a function? And is v the same thing as V? If so, then the equation is nonlinear in v, which doesn't have a general method to solve, although maybe its possible...
     
  6. Sep 23, 2011 #5

    rcgldr

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    [tex]\frac{dv}{v^2 + 9.8 / c}= -c \ dt[/tex]

    This is similar to the formula used for free fall, wiki article:

    wiki_free_fall.htm
     
    Last edited: Sep 23, 2011
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