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Firing a spherical bullet into a watertank

  • Thread starter TheMan112
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  • #1
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[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.
 

Answers and Replies

  • #2
Alfi
Horizontally ?

Or perpendicular to a smooth water surface?
 
  • #3
tiny-tim
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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.
Hi TheMan112! :smile:

Hint: a = dv/dt. :smile:
 
  • #4
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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:
  • #5
43
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Horizontally ?

Or perpendicular to a smooth water surface?
Imagine we mounted the muzzle of a gun to a hole in the side of the watertank, fireing into the water horizontally.

Hi TheMan112! :smile:

Hint: a = dv/dt. :smile:
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.
 
  • #6
tiny-tim
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[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]
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:
 
  • #7
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Seems like that did it, thanks!
 

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