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- Thread starter Big-Daddy
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tiny-tim

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Hi Big-Daddy!

Tell us what*you* think, and why, and then we'll comment!

Tell us what

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For free-fall I could model it pretty easily:

[tex]mg - kv^2 = m \cdot \frac{dv}{dt}[/tex]

But this doesn't seem to be remotely of the same difficulty. Velocity in free fall is always in the same direction as acceleration, but in the projectile motion case, the velocity is defined by the initial projection angle and velocity (which would be given of course, along with the values of m, g and k).

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Nugatory

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but in the projectile motion case, the velocity is defined by the initial projection angle and velocity (which would be given of course, along with the values of m, g and k).

Try writing the differential equation in terms of vectors [itex]\vec{v}[/itex] and [itex]\vec{F}=m\vec{a}[/itex]. That will at least get the problem modeled properly, and you can decompose it into coupled differential equations for ##x(t)## and ##y(t)##.

The initial angle and speed provide the boundary conditions you'll need to determine the arbitrary constants that show up in the solutions of the differential equations.

(And I have to caution you that solving these equations is a non-trivial problem).

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Try writing the differential equation in terms of vectors [itex]\vec{v}[/itex] and [itex]\vec{F}=m\vec{a}[/itex]. That will at least get the problem modeled properly, and you can decompose it into coupled differential equations for ##x(t)## and ##y(t)##.

Well ok, perhaps this is an analogous equation to mine for free-fall, where v has now been replaced by a vector?

Question is, how to decompose this into my x(t) and y(t) differential equations?

- #6

jhae2.718

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Question is, how to decompose this into my x(t) and y(t) differential equations?

Equate the components of the acceleration and force vectors to obtain the scalar equations of motion.

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tiny-tim

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(just got up :zzz:)

For free-fall I could model it pretty easily:

[tex]mg - kv^2 = m \cdot \frac{dv}{dt}[/tex]

ok, same, but with vectors …

[itex]-mg\mathbf{y} - kf(\mathbf{v})\mathbf{v} = m \cdot \frac{d\mathbf{v}}{dt}[/itex]

where ##f(\mathbf{v})## = … ?

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Equate the components of the acceleration and force vectors

What do you mean?

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ok, same, but with vectors …

[itex]-mg\mathbf{y} - kf(\mathbf{v})\mathbf{v} = m \cdot \frac{d\mathbf{v}}{dt}[/itex]

where ##f(\mathbf{v})## = … ?

I'm not sure I understand what that y is doing there. As for f(v), not sure ... looks like it should just be v to me, but I'm not sure why you wrote it as such then ...

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arildno

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f is a scalar, non-negative function.

What ought f to be then?

What ought f to be then?

- #11

tiny-tim

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I'm not sure I understand what that y is doing there.

because gravity is mg directly downwards,

so the vector for the force of gravity is -mg in the y direction, ie -mg time the unit vector in the y direction, ie -mg

As for f(v), not sure ... looks like it should just be v to me

yes, but you need to write it in terms of the vector

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Equate the components of the acceleration and force vectors to obtain the scalar equations of motion.

jhae meansWhat do you mean?

As for the original question:

The air resistance force has a magnitude

Do the same for the

And then you'll have the force components due to air resistance to use in the equations relating

p.s. This is worth repeating:

(And I have to caution you that solving these equations is a non-trivial problem).

- #13

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So far I have

[tex]F_x = m \cdot \frac{d^2x}{dt^2} = -k \cdot \frac{dx}{dt}[/tex]

and

[tex]F_y = m \cdot \frac{d^2x}{dt^2} = -mg \cdot cos(\theta) - k \cdot \frac{dx}{dt}[/tex]

I think there is a problem in how I have resolved the weight though. How can I do this better?

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A couple of problems to clear up in your equations:

1. The weight is -

2. Also, you have incorrectly multiplied

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arildno

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1. The weight is -mg. There is no cosine term involved in the weight, since it always acts downward, in the -y-direction, regardless of the angle of the trajectory.

Ok then so maybe

[tex]F_y = m \cdot \frac{d^2y}{dt^2} = -mg - k \cdot \frac{dy}{dt}[/tex]

2. Also, you have incorrectly multipliedkv^{2}and -v/_{x}v-- so you are missing a factor ofvin the air resistance expression. (And the same error occurs in they-equation).

I'm not sure I understand ... by v, do you mean the resultant of v

[tex]F_y = m \cdot \frac{d^2y}{dt^2} = -mg - k \cdot \frac{dy}{dt} \cdot ((\frac{dy}{dt})^2+(\frac{dx}{dt})^2)^{1/2}[/tex]

and

[tex]F_x = m \cdot \frac{d^2x}{dt^2} = - k \cdot \frac{dy}{dt} \cdot ((\frac{dy}{dt})^2+(\frac{dx}{dt})^2)^{1/2}[/tex]

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arildno

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In your last last line, the "dy/dt" outside the root should be replaced with "dx/dt"

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Yes to both.I'm not sure I understand ... by v, do you mean the resultant of v_{x}and v_{y}, and by v_{x}you mean dx/dt?

Apart from arildno's correction, you got it.

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And the boundary conditions would involve me specifying initial y and x displacement from the origin as well as initial x velocity and y velocity (which can be found as the cos and sin components respectively of the initial total magnitude of velocity, making sure the y initial velocity is positive if the point is travelling upwards initially and negative if it is travelling downwards initially), and nothing else?

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Nugatory

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And the boundary conditions would involve me specifying initial y and x displacement from the origin as well as initial x velocity and y velocity (which can be found as the cos and sin components respectively of the initial total magnitude of velocity, making sure the y initial velocity is positive if the point is travelling upwards initially and negative if it is travelling downwards initially), and nothing else?

Yes.

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