Equation of Motion for a Projectile Under Quadratic Air Resistance

[Oblio]

I realized that after I wrote it.

so subbing u back in I get...

-sqrt[m/gc]arctan [sqrt[c/mg] xv

 

[learningphysics]

I realized that after I wrote it.

so subbing u back in I get...

-sqrt[m/gc]arctan [sqrt[c/mg] xv

yes...but you need a constant so:

$$t = -\sqrt{\frac{m}{gc}}arctan(\sqrt{\frac{c}{mg}}*v) + C$$

we can solve for C by subsituting in t = 0, v = v0

so $$C = \sqrt{\frac{m}{gc}}arctan(\sqrt{\frac{c}{mg}}*v0)$$

So $$t = -\sqrt{\frac{m}{gc}}arctan(\sqrt{\frac{c}{mg}}*v) + \sqrt{\frac{m}{gc}}arctan(\sqrt{\frac{c}{mg}}*v0)$$

 

[Oblio]

Ok, and somehow my two square roots will boil down to v(ter)/g and vo/v(ter) ?

 

[learningphysics]

EDIT:

actually no... you don't need to do this... sorry about that...

you can solve for vter... at the terminal velocity the net force is 0...

mg = cv^2

vter = sqrt(mg/c)

that should work...

 

[Oblio]

I only need the time to the top of the trajectory though?
Are you sure I need to find the time for down?

 

[learningphysics]

I only need the time to the top of the trajectory though?
Are you sure I need to find the time for down?

yeah you're right. you don't need that part. I think you're almost done.

 

[Oblio]

yeah you're right. you don't need that part. I think you're almost done.

lol phew I was worried for a sec.

have I already defined the relation that will give me

t(top) = $$\frac{v(ter)}{g}$$arctan($$\frac{vo}{v(ter)}$$) ?

 

[learningphysics]

lol phew I was worried for a sec.

have I already defined the relation that will give me

t(top) = $$\frac{v(ter)}{g}$$arctan($$\frac{vo}{v(ter)}$$) ?

yeah, look at post #92, with the constant evaluated... you want the time when v = 0. What time do you get? Try to sub in vter...

 

[Oblio]

Ok, one sec. But vter is when v=0 ?

 

[Oblio]

just C?

 

[learningphysics]

Ok, one sec. But vter is when v=0 ?

No. look at post #94 for vter.

 

[learningphysics]

Ok, one sec. But vter is when v=0 ?

v=0 is at the top of the trajectory.

 

[Oblio]

No. look at post #94 for vter.

Ok I thought you were saying that :P

 

[Oblio]

I got t=c for v=0

 

[Oblio]

and vter = sqrt[mg/c]

 

[learningphysics]

I got t=c for v=0

yes. And I've calculated C in post #92.

 

[Oblio]

I can manipulate the right to get the correct v/vter, but not the left yet. I need g in the denominator, but I also need it for vter...

 

[learningphysics]

I can manipulate the right to get the correct v/vter, but not the left yet. I need g in the denominator, but I also need it for vter...

vter = sqrt(mg/c)

what is vter/g ?

 

[Oblio]

sqrt[m/c]

but now with all the stuff C brought in I don't have the equation I'm after anymore

 

[learningphysics]

sqrt[m/c]

No, vter/g = sqrt(m/cg)

but now with all the stuff C brought in I don't have the equation I'm after anymore

I don't understand... you need to get the time to the top of the trajectory right?

time to the top of the trajectory = C.

 

[Oblio]

lol phew I was worried for a sec.

have I already defined the relation that will give me

t(top) = $$\frac{v(ter)}{g}$$arctan($$\frac{vo}{v(ter)}$$) ?

Yeah, the question said to solve it to this:

which I have withOUT c.. lol.. I don't get it

 

[Oblio]

it's the exact same thing though..
whats that mean?

 

[learningphysics]

it's the exact same thing though..
whats that mean?

$$t = -\sqrt{\frac{m}{gc}}arctan(\sqrt{\frac{c}{mg}}*v) + \sqrt{\frac{m}{gc}}arctan(\sqrt{\frac{c}{mg}}*v0)$$

you need t when v = 0,

$$t_{top} = \sqrt{\frac{m}{gc}}arctan(\sqrt{\frac{c}{mg}}*v0)$$

And this equals: $$\frac{vter}{g}arctan(\frac{v0}{vter})$$

 

[Oblio]

Your right. I get it!

 

[Oblio]

when you do

sqrt[mg/c] /g

=sqrt[mg/c] x 1/g
=sqrt[mg/cg]

Now I know this is wrong, but...why?

 

[learningphysics]

when you do

sqrt[mg/c] /g

=sqrt[mg/c] x 1/g
=sqrt[mg/cg]

Now I know this is wrong, but...why?

when you bring the g under the square root... you need to square it... g = sqrt(g^2)

so your last step would be

=sqrt[mg/(cg^2)]

 

[Oblio]

*high school flash back*

Right!

 

[Oblio]

with gravity, wouldn't this give me negative time values?