Equation of Motion for a Projectile Under Quadratic Air Resistance

[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.