Equation of Motion for a Projectile Under Quadratic Air Resistance

[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 ?

 

[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.

 

[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]

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.

 

[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]

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)

 

[weatherhead]

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]

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]

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

which equation?

 

[Oblio]

v=sqrt[ma +mg/-c]

 

[learningphysics]

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]

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]

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)

 

[learningphysics]

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

Yes.

 

[Oblio]

so will it be:

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

 

[learningphysics]

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:

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

 

[Oblio]

So, just considering you, integrating that would give me ut +C?

 

[Oblio]

I'm not sure if that's how you meant me to use a substitution.

 

[Oblio]

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...

lol in addition, where did dt come from?
Were you using my original equation of v= sqrt[ma+mg/c] ?

 

[learningphysics]

lol in addition, where did dt come from?
Were you using my original equation of v= sqrt[ma+mg/c] ?

ma= -mg - cv^2 for the upwards part...

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

I get the dt from above.

when you rearrange this equation you get

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

 

[Oblio]

Right. I always forget to put dt in the denominator, and that you can't just make a into 'dv' lol.

Ok, am I kind of using your substitution correctly above?

 

[learningphysics]

I'm not sure if that's how you meant me to use a substitution.

It's a method of solving integrals, called integration by substitution.

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

$$du = \sqrt{\frac{c}{mg}}*dv$$

solve for v and dv in the above equations, and substitute into the integral:

tf-ti = $$\frac{-m}{mg}*\frac{1}{\sqrt{\frac{c}{mg}}}\int\frac{du}{1+u^2}$$

now you can use the arctan for the right side...

once you get the integral in terms of u... you can substitute $$\sqrt{\frac{c}{mg}}*v$$ for u

 

[Oblio]

It's a method of solving integrals, called integration by substitution.

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

$$du = \sqrt{\frac{c}{mg}}*dv$$

Not dv/dt?

 

[Oblio]

It's a method of solving integrals, called integration by substitution.

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

$$du = \sqrt{\frac{c}{mg}}*dv$$

solve for v and dv in the above equations, and substitute into the integral:

tf-ti = $$\frac{-m}{mg}*\frac{1}{\sqrt{\frac{c}{mg}}}\int\frac{du}{1+u^2}$$

now you can use the arctan for the right side...

once you get the integral in terms of u... you can substitute $$\sqrt{\frac{c}{mg}}*v$$ for u

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

Bring out the constants $$\frac{-m}{mg}$$


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

the numerator and denominator are factored out above...


Am I correct in say that $$\sqrt{\frac{c}{mg}}$$ in the numerator and denominator in the integral cancel out leaving just v and dv?

I'm not following the step in getting $$\frac{1}{\sqrt{\frac{c}{mg}}}$$ outside...

 

[learningphysics]

Do you have a calculus textbook handy by any chance?

 

[Oblio]

at home... not here at school. Sorry. Why?