Equation of Motion for a Projectile Under Quadratic Air Resistance

[learningphysics]

Ok, got it. It's more of a notation.

Yeah, for the most part...

Yeah I follow your online example now.

cool.

 

[Oblio]

we not only need to substitute u... we also need to take care of the dv, and get du instead...

so:

du = \sqrt{c/mg}*dv

dv = \frac{du}{\sqrt{c/mg}}

So substitute that into the integral for dv...

what do you get for the integral now?

\frac{-m}{mg} \int \frac{\frac{du}{sqrt[c/mg]}}{1 + c/mg* v^2}

 

[learningphysics]

\frac{-m}{mg} \int \frac{\frac{du}{sqrt[c/mg]}}{1 + c/mg* v^2}

yes, but you also need the u substitution...

 

[Oblio]

You mean du?

 

[learningphysics]

You mean du?

you've taken care of the du... but you still have v in the integral... you want u... we had it before

1+u^2

 

[Oblio]

\frac{-m}{mg} \int \frac{\frac{du}{sqrt[c/mg]}}{1 + c/mg* v^2}

right...

\frac{-m}{mg} \int \frac{\frac{du}{sqrt[c/mg]}}{1+ u^2}

 

[learningphysics]

right...

\frac{-m}{mg} \int \frac{\frac{du}{sqrt[c/mg]}}{1+ u^2}

exactly... now take all the constants outside the integral... and we get:

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

now we can apply arctan...

 

[Oblio]

Do you treat the integral sign almost as an equality when moving things in and out?

 

[learningphysics]

Do you treat the integral sign almost as an equality when moving things in and out?

not like an equal sign... more like a parentheses...

ie: 4x^2y+8xy = 4(x^2y + 2xy) = 4xy(x+2)

it's the same way I'm moving out the constants...

 

[Oblio]

Ok because I am having trouble matching what you got taking out the last constants

 

[Oblio]

what step did you take?

 

[learningphysics]

Ok because I am having trouble matching what you got taking out the last constants

show me what you get... I might have made a mistake.

 

[Oblio]

One more question first, how do we treat it like a parantheses, when you don't really have anything outside the 'bracket' to multiply or divide, in the simplification process.

I tried lots of things, but logically I'm not sure the proper logic behind taking out those constants when there's no 'other side' to multiply/divide etc...
know what i mean?

 

[learningphysics]

One more question first, how do we treat it like a parantheses, when you don't really have anything outside the 'bracket' to multiply or divide, in the simplification process.

I tried lots of things, but logically I'm not sure the proper logic behind taking out those constants when there's no 'other side' to multiply/divide etc...
know what i mean?

I'm not sure I understand... for example:

\int5xdx = 5\int xdx

For any integral \int A*f(x)dx = A\int f(x)dx where A is a constant.

we're just factoring it out... so you don't need an other side...

never mind about taking the constants out...

suppose I had it like this:

\int\frac{-m}{mg}*\frac{\frac{du}{sqrt[c/mg]}}{1+ u^2}

now, I just want to clean this up a little... simplify it... don't take anything outsdie the integral.

 

[Oblio]

You mean for me to to it?

I know and understand your first rule there, but does that apply to a multi layer division?

 

[learningphysics]

You mean for me to to it?

Yeah.

I know and understand your first rule there, but does that apply to a multi layer division?

Which rule?

 

[Oblio]

\int5xdx = 5\int xdx

For any integral \int A*f(x)dx = A\int f(x)dx where A is a constant.

QUOTE]

 

[learningphysics]

\int5xdx = 5\int xdx

For any integral \int A*f(x)dx = A\int f(x)dx where A is a constant.

QUOTE]

don't worry about taking anything out of the integral... just simplify it inside the integral.

 

[Oblio]

Yeah, that's what I'm aiming for, but when I simplify equations; mentally I do the 'what you do to one side, do to the other' etc. With nothing outside of the integral, how do we bring out sqrt[c/mg] ?

 

[learningphysics]

Yeah, that's what I'm aiming for, but when I simplify equations; mentally I do the 'what you do to one side, do to the other' etc. With nothing outside of the integral, how do we bring out sqrt[c/mg] ?

don't bring it out:

\int\frac{-m}{mg}*\frac{\frac{du}{sqrt[c/mg]}}{1+ u^2}

this equals

\int\frac{-m}{mg}*\frac{1}{sqrt[c/mg]}*\frac{du}{1+ u^2}

now, the first thing you can do is cancel the m's in the numerator and denominator... what else can you do to simplify...

 

[Oblio]

Besides moving it out of the integral?

 

[learningphysics]

Besides moving it out of the integral?

yes. forget about moving it out of the integral...

 

[Oblio]

yes. forget about moving it out of the integral...

we have 2 fractions multiplied together...


\frac{-1}{g} x \frac{1}{/sqrt{c/mg}}

I would think the sqrt would need to removed maybe...

 

[learningphysics]

we have 2 fractions multiplied together...


\frac{-1}{g} x \frac{1}{/sqrt{c/mg}}

I would think the sqrt would need to removed maybe...

can you do anything with the g that is there and the g inside the square root?

also try to clean up the fraction so that I don't have additional fractions inside the numerator and denominator... ie: in the numerator I want stuff being multiplied... in the denominator I want stuff being multiplied...

 

[Oblio]

I'm sure this isn't hard but I can't see anything...

 

[Oblio]

I know the fractions are technically the same thing as -1/g/sqrt[c/mg], but that doesn't help

 

[learningphysics]

I know the fractions are technically the same thing as -1/g/sqrt[c/mg], but that doesn't 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]

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 that's wrong...

 

[learningphysics]

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

is it close to...
sqrt[c/(msqrt[g])] lol that's 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]

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]

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 isn't it?

 

[learningphysics]

That IS arctan isn't it?

yeah.

 

[Oblio]

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

 

[learningphysics]

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]

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?