What is the maths behind projectiles (including air resistance)?

[Georgepowell]

E.g. A football of mass 1kg is thrown with a speed of 10m/s at an angle of 30 degrees from a horizontal plane. Take air resistance (in Newtons) as 1/15 th of the speed of the ball in the exact opposite direction that the ball is traveling. How long is the ball in the air? and how far (horizontally) has it traveled at the point of landing?

I understand how to figure it out if I can ignore air resistance, but how would I answer this question?

I am not so much interested in the answer, but more how to figure out the answer, I have just given you that example so it gives you something to answer if you need an example.

 

[atyy]

Stokes' Law
http://www.phy.uct.ac.za/courses/phy300w/millikan/

Correction to Stokes' Law
http://www.sigmaxi.org/meetings/archive/forum.2000.millikan.shtml
http://scienceweek.com/2004/rmps-27.htm

Much more than a correction to Stokes's Law:
http://ocw.mit.edu/OcwWeb/Aeronautics-and-Astronautics/16-100Fall-2005/LectureNotes/index.htm

 

[Georgepowell]

That tells me how to figure out the air resistance, and not how to predict its movement after I know it.

Anybody else?

 

[Hootenanny]

That tells me how to figure out the air resistance, and not how to predict its movement after I know it.

Anybody else?

One would need to formulate a system of differential equations using Newton's second law. The exercise is fairly straight forward for the case of linear drag (as you have described in your opening post).

 

[Georgepowell]

Ok, so from what I already know I can write the instantaneous acceleration in terms of the velocity at that time, but I can't seem to figure out how to have acceleration/velocity as a function of time.

Any help on this?

 

[Hootenanny]

Ok, so from what I already know I can write the instantaneous acceleration in terms of the velocity at that time, but I can't seem to figure out how to have acceleration/velocity as a function of time.

Any help on this?

We have Newton's second law:

\mathbf{F} = m\frac{d^2\mathbf{r}}{dt^2}

Where r = xi + yj is position, where x=x(t) and y=y(t). Splitting into vertical and horizontal components:

-mg -\kappa y^\prime = my^{\prime\prime}

-\kappa x^\prime = mx^{\prime\prime}

Where \kappa = 1/15. In canonical form:

y^{\prime\prime} + \frac{\kappa}{m} y^\prime = g

x^{\prime\prime} + \frac{\kappa}{m} x^\prime = 0

Both of the above are second order differential equations with constant coefficients. Both are separable and straightforward to solve.

 

[Georgepowell]

https://www.physicsforums.com/showthread.php?t=302852

That is the answer that I was looking for (and found after doing integration and differentiation at school).