1. Sep 23, 2007

Oblio

I have to write the equation of motion for a projectile thrown vertically, under quadratic air resistance.

The forces acting on the projectile then are, gravity and the quadratic air resistance.

Now, both forces are opposing the projectile, would that mean that if I make gravity negative (-mg) that my air resistance would also need to be negative?

ma = -mg-cv^2 ?

2. Sep 23, 2007

mgb_phys

Air resistance always acts to slow the object, so on the way up it works in the same direction as gravity - pulling the object down.
But on the way back down it acts agaisnt gravity, slowing the descent.

3. Sep 23, 2007

Oblio

Ok good.
I'm starting with the upward motion so they're both negative.

Now, I need to find that the time to the top of the trajectory is

t= (v$$_{ter}$$/g) arctan(vo/v$$_{ter}$$)

I think I would need to work in that switch where the quadratic air resistance becomes positive? Am I on the right track?

I solved v to be
v=sqrt[ma +mg/-c]

4. Sep 23, 2007

Oblio

I am completely failing to see how trig is finding its way into this, and I was supposed to find the v as a function of t, and technically time is found deep within 'c', do you think my equation is satisfactory as a function of t?

5. Sep 23, 2007

learningphysics

Is it thrown straight up?

So upwards... ma = -mg-cv^2

downwards... ma = -mg+cv^2

You need to solve the differential equation

mdv/dt = -mg - cv^2 for the upwards part...

[/quote]

I get the dt from above.

Last edited: Sep 24, 2007
6. Sep 23, 2007

Oblio

That's where I'm failing to understand it.. why would arctan be needed?

7. Sep 23, 2007

Oblio

p.s. I AM trying to upgrade my lacking integration skills...

8. Sep 23, 2007

learningphysics

Personally, I had to look up that integral... so that's why I know arctan works.

9. Sep 23, 2007

Oblio

I'm not sure HOW any trig would find its way into this integral though..

(hehe thanks though)

10. Sep 23, 2007

trig pops up in the strangest of places, and if you see it in a question like this where it looks like it wouldn't have any relevance, you should be able to smell the calculus. You will learn to recognise these

11. Sep 23, 2007

learningphysics

Yeah, I understand.

But all you need is to know that $$arctan(x) = \int{\frac{1}{1+x^2}dx}$$

12. Sep 23, 2007

Oblio

before I go to that,is my equation a satisfactory function of time?

13. Sep 23, 2007

Oblio

because c = gamma(D), and D='something with time in it, lol I forget) I'm not sure if it's good enough.

14. Sep 23, 2007

learningphysics

which equation?

15. Sep 23, 2007

Oblio

v=sqrt[ma +mg/-c]

16. Sep 23, 2007

learningphysics

that equation is correct, but I'm not sure how you want to use it... You need to do the integral I posted:

dt = -mdv/(mg+cv^2)

integrate both sides of this...

17. Sep 23, 2007

Oblio

The question said to solve it for v as a function of t though.
Do you mean to do that afterwards?

18. Sep 23, 2007

learningphysics

yes.

19. Sep 23, 2007

Oblio

Ok.
I think these questions were written so that I do the order prescribed, whichis actually wrong:P

20. Sep 23, 2007

Oblio

c is a complicated number.. not sure how to integrate it..

21. Sep 23, 2007

learningphysics

It's just a constant. you may need to do a variable substitution...

dt = -mdv/(mg+cv^2)

tf - ti = $$\int\frac{-mdv}{mg(1+(c/mg)v^2}$$

tf - ti = $$\frac{-m}{mg}\int\frac{dv}{1+(c/mg)v^2}$$

22. Sep 23, 2007

Oblio

this entire section's a constant.. yes? : (c/mg)

23. Sep 23, 2007

learningphysics

Yes.

24. Sep 23, 2007

Oblio

so will it be:

v / [(c/mg)(v^3/3) + v^2 ] ?

25. Sep 24, 2007

learningphysics

No... try to get the integral to look like the integral I posted for arctan...

use a variable substitution:

$$u = \sqrt{\frac{c}{mg}}*v$$