Firing a spherical bullet into a watertank

[TheMan112]

[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:

$$F = -k \cdot v$$ where k is a constant.

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

$$a = \frac{F}{m} = - {\frac{k v}{m}}$$

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 don't know how to do it without getting a recursive expression.

 

[Alfi]

Horizontally ?

Or perpendicular to a smooth water surface?

 

[tiny-tim]

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 don't know how to do it without getting a recursive expression.

Hi TheMan112!

Hint: a = dv/dt.

 

[Lojzek]

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.

 

[TheMan112]

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!

Hint: a = dv/dt.

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

$$m\ddot{x}-k\dot{x}=0$$

...I get:

$$\frac{dx}{dt} = e^{\frac{m}{k} t}$$

Which in turn gives:

$$s = \int_{0}^{\infty} v(t) dt = \int_{0}^{\infty} e^{\frac{m}{k} t} dt = -\frac{m}{k}$$

Which gives an answer wrong by a factor of 10^3.

 

[tiny-tim]

$$\frac{dx}{dt} = e^{\frac{m}{k} t}$$

Which in turn gives:

$$s = \int_{0}^{\infty} v(t) dt = \int_{0}^{\infty} e^{\frac{m}{k} t} dt = -\frac{m}{k}$$

Hi TheMan112!

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

It should be:

$$\frac{dx}{dt} = v_0 e^{\frac{-k}{m} t}$$

Which in turn gives:

$$s = \int_{0}^{\infty} v(t) dt = v_0 \int_{0}^{\infty} e^{\frac{-k}{m} t} dt\,=\,...\,?$$

 

[TheMan112]

Seems like that did it, thanks!