Equation of Motion for a Projectile Under Quadratic Air Resistance

[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?