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Firing a spherical bullet into a watertank

  1. Apr 30, 2008 #1
    [SOLVED] Firing a spherical bullet into a watertank

    I've got a problem, involving non-constant acceleration:

    If we fire a spherical bullet horizontally into a watertank, how far will the bullet traverse?

    I've figured as much that a spherical bullet provides a retarding force:

    [tex]F = -k \cdot v[/tex] where k is a constant.

    This should provide the following non-constant acceleration due to Newtons 2nd law.

    [tex]a = \frac{F}{m} = - {\frac{k v}{m}}[/tex]

    I'm thinking I should integrate two times over a(t) to get an expression for x(t), but since "a" is proportional to v(t) and not directly to t, I dont know how to do it without getting a recursive expression.
     
  2. jcsd
  3. Apr 30, 2008 #2
    Horizontally ?

    Or perpendicular to a smooth water surface?
     
  4. Apr 30, 2008 #3

    tiny-tim

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    Hi TheMan112! :smile:

    Hint: a = dv/dt. :smile:
     
  5. Apr 30, 2008 #4
    You have an equation with separable variables (the form f(y)dy=g(x)dx). This can be solved by integrating both sides: left from y1 to y2, right from x1 to x2.
    Another option is solving for v(x):

    dv/dx=dv/dt*dt/dx=(dv/dt)/v

    After you get v(x) you can get corresponding t from dt=dx/v.

    But I don't think that bullets experience linear drag. At high speeds F=-kv^2 is probably a better apoximation.
     
    Last edited: Apr 30, 2008
  6. May 2, 2008 #5
    Imagine we mounted the muzzle of a gun to a hole in the side of the watertank, fireing into the water horizontally.

    Haha, I know that, and from the following differential equation...

    [tex]m\ddot{x}-k\dot{x}=0[/tex]

    ...I get:

    [tex]\frac{dx}{dt} = e^{\frac{m}{k} t}[/tex]

    Which in turn gives:

    [tex]s = \int_{0}^{\infty} v(t) dt = \int_{0}^{\infty} e^{\frac{m}{k} t} dt = -\frac{m}{k}[/tex]

    Which gives an answer wrong by a factor of 10^3.
     
  7. May 2, 2008 #6

    tiny-tim

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    Hi TheMan112! :smile:

    Somehow you've managed to get a nearly right answer with the wrong reasoning. :rolleyes:

    It should be:

    [tex]\frac{dx}{dt} = v_0 e^{\frac{-k}{m} t}[/tex]

    Which in turn gives:

    [tex]s = \int_{0}^{\infty} v(t) dt = v_0 \int_{0}^{\infty} e^{\frac{-k}{m} t} dt\,=\,...\,?[/tex] :smile:
     
  8. May 3, 2008 #7
    Seems like that did it, thanks!
     
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