#### 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 ?

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#### mgb_phys

Homework Helper
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.

#### 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]

#### 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?

#### learningphysics

Homework Helper
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.

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#### Oblio

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

#### Oblio

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

#### learningphysics

Homework Helper
p.s. I AM trying to upgrade my lacking integration skills...
Personally, I had to look up that integral... so that's why I know arctan works. #### Oblio

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

(hehe thanks though)

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

#### learningphysics

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

(hehe thanks though)
Yeah, I understand.

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

#### Oblio

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

#### Oblio

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

#### learningphysics

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

#### Oblio

v=sqrt[ma +mg/-c]

#### learningphysics

Homework Helper
v=sqrt[ma +mg/-c]
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...

#### Oblio

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

#### learningphysics

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

#### Oblio

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

#### Oblio

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

#### learningphysics

Homework Helper
c is a complicated number.. not sure how to integrate it..
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 = $$\frac{-m}{mg}\int\frac{dv}{1+(c/mg)v^2}$$

#### Oblio

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

Homework Helper

#### Oblio

so will it be:

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

#### learningphysics

Homework Helper
so will it be:

v / [(c/mg)(v^3/3) + v^2 ] ?
No... try to get the integral to look like the integral I posted for arctan...

use a variable substitution:

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