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Drag equation and d(kinetic energy)/d(displacement)

  1. Feb 2, 2016 #1
    1. The problem statement, all variables and given/known data

    Given the drag equation and a projectile's initial kinetic energy and mass, find the distance in which it will come to rest (or at least, become absurdly slow).

    2. Relevant equations

    https://en.wikipedia.org/wiki/Drag_equation

    3. The attempt at a solution

    Force = .5 * fluid density * drag coefficient * area * velocity^2.
    Velocity = sqrt(2 * energy / mass)
    Force = fluid density * drag coefficient * area * energy / mass.

    When the bullet emits dE energy, it travels dE/F.

    Thus, position delta = dE/dP = dE/F = dE / (fluid density * drag coefficient * area * energy at step / mass)

    dE/dP = (mass / (fluid density * drag coefficient * area)) 1/E dE

    Everything in front of 1/E is a constant, so it isn't integrated. The integral of 1/E is ln(E), so the solution is

    displacement = (mass / (fluid density * drag coefficient * area)) (log(E0) - log(1 eV))

    4. ???

    WTF is ln(E)?
     
  2. jcsd
  3. Feb 2, 2016 #2
    Your notation is a mystery.
     
  4. Feb 2, 2016 #3
    What in particular? I just took the drag equation and rearranged it. I want to figure out how far the projectile moves based on its mass and speed. How would you do it?
     
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