#### Oblio

I know the fractions are technically the same thing as -1/g/sqrt[c/mg], but that doesnt help

#### learningphysics

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I know the fractions are technically the same thing as -1/g/sqrt[c/mg], but that doesnt help
what is $$\frac{g}{\sqrt{g}}$$?

#### Oblio

In trying things out manually I'm finding that its sqrt[g] but i definitely didnt know that before now... hmm

#### learningphysics

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In trying things out manually I'm finding that its sqrt[g] but i definitely didnt know that before now... hmm
exactly... can you further simplify the integral?

#### Oblio

Its hard since its in the denominator in the square root...

is it close to...
sqrt[c/(msqrt[g])] lol thats wrong...

#### learningphysics

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Its hard since its in the denominator in the square root...

is it close to...
sqrt[c/(msqrt[g])] lol thats wrong...
$$\frac{-1}{g}\times\frac{1}{\sqrt{c/mg}}$$

=

$$\frac{-1}{\sqrt{gc/m}}$$

=

$$-\sqrt{\frac{m}{gc}}$$

#### Oblio

I'll have to do some research on that, I can't say I'm following manipulating such distant numbers..

#### Oblio

$$\frac{-1}{g}\times\frac{1}{\sqrt{c/mg}}$$

=

$$\frac{-1}{\sqrt{gc/m}}$$

=

$$-\sqrt{\frac{m}{gc}}$$
For the time being, I'm left with

$$-\sqrt{\frac{m}{gc}}$$ $$\int$$ $$\frac{du}{1+u^2}$$

Now integrate it I assume

#### learningphysics

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For the time being, I'm left with

$$-\sqrt{\frac{m}{gc}}$$ $$\int$$ $$\frac{du}{1+u^2}$$

Now integrate it I assume
yeah, use arctan. we should actually have limits on that integral... so:

$$-\sqrt{\frac{m}{gc}}$$ $$\int_{u_{initial}}^{u_{final}}$$ $$\frac{du}{1+u^2}$$

#### Oblio

yeah, use arctan. we should actually have limits on that integral... so:

$$-\sqrt{\frac{m}{gc}}$$ $$\int_{u_{initial}}^{u_{final}}$$ $$\frac{du}{1+u^2}$$

We often didn't use limits. Why this time?

#### learningphysics

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We often didn't use limits. Why this time?
Yeah, we don't need the limits... but we should add a constant when we take the integral...

#### Oblio

$$\int_{u_{initial}}^{u_{final}}$$ $$\frac{du}{1+u^2}$$
That IS arctan isnt it?

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#### Oblio

Ok so if arctan = that, I need to find out what the integral of arctan is..

#### learningphysics

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Ok so if arctan = that, I need to find out what the integral of arctan is..
no arctan IS the integral.

#### Oblio

I realized that after I wrote it.

so subbing u back in I get...

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

#### learningphysics

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

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#### Oblio

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

#### learningphysics

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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...

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#### 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

Homework Helper
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

Homework Helper
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 ?

just C?