Trajectory: integrating for position

[Oblio]

well, my c = vsin(theta)

 

[learningphysics]

well, my c = vsin(theta)

exactly.

vy = -gt + vsin(theta)
y = (-1/2)gt^2 + (vsin(theta))t + D

do, the same type of thing for D using the position equation... at t = 0, y = 0...

 

[Oblio]

D=0... ?

 

[Oblio]

technically my equations were right weren't they? I just wasnt solving for t=0 yet.

 

[learningphysics]

D=0... ?

yes.

vy = -gt + vsin(theta)
y = (-1/2)gt^2 + (vsin(theta))t

now, we could have done this directly... use the equations for displacement with accleretated motion ie:

d = v0*t + (1/2)at^2

but since the question asked to integrate... we did it this way...

can you do the same type of thing in the x direction?

$$a_x = 0$$

 

[Oblio]

It's all necessary with no horizontal forces?

 

[Oblio]

0 = ma

C=v(x)
v(x) =vcos(theta), C=Vcos(theta)

y=v(x)t +D
y=D

at t=0, D=0 also.

 

[learningphysics]

0 = ma

C=v(x)
v(x) =vcos(theta), C=Vcos(theta)

y=v(x)t +D
y=D

at t=0, D=0 also.

Yes, so:

x = vcos(theta)t
y = vsin(theta)t - (1/2)gt^2

you can eliminate t from these two equations now.

 

[Oblio]

algebraically or through another situation such as (t=0) (not that one I realize though)

 

[learningphysics]

algebraically or through another situation such as (t=0) (not that one I realize though)

algebraically.

 

[Oblio]

set them equal?

 

[Oblio]

nope. lol

 

[learningphysics]

nope. lol

solve for t in one... substitute that into the other.

 

[Oblio]

ok, it asks for the trajectory as a function of x so:

y=vsin(theta)x/cos(theta) - (1/2)g x^2/v^2cos(theta)^2

 

[learningphysics]

ok, it asks for the trajectory as a function of x so:

y=vsin(theta)x/cos(theta) - (1/2)g x^2/v^2cos(theta)^2

there should be vcos(theta) in the denominator in the first term.

 

[Oblio]

I think that's right according to what I'm supposed to be comparing it too...

 

[Oblio]

there should be vcos(theta) in the denominator in the first term.

sorry.
had that in ink ( i promise!)

 

[learningphysics]

sorry.
had that in ink ( i promise!)

lol! I believe you. So the v's will cancel in that first term. you get xtan(theta)

 

[Oblio]

Yup!

 

[learningphysics]

So now you have to get the path again, but this time with air resistance?

 

[Oblio]

I actually did that FIRST, thinking I was doing part a, but I actually did b.

I think this question is basically done, I can show you my other section maybe?
They do basically match.

 

[learningphysics]

I actually did that FIRST, thinking I was doing part a, but I actually did b.

I think this question is basically done, I can show you my other section maybe?
They do basically match.

Sure... go ahead and show it.

 

[Oblio]

y= (v$$_{yo}$$ + v$$_{ter}$$x) / v$$_{xo}$$ + v$$_{ter}$$$$\tau$$ ln (1 - x / v$$_{xo}$$$$\tau$$)

I'll skip a simplification step unless its needed: Used taylor's series which allowed cancellations.

y=v$$_{yo}$$x / v$$_{xo}$$ - x^2v$$_{ter}$$$$\tau$$ / 2v$$_{xo}$$$$^{2}$$$$\tau$$$$^{2}$$

 

[Oblio]

as usual subscripts are superscripts...

 

[learningphysics]

y= (v$$_{yo}$$ + v$$_{ter}$$x) / v$$_{xo}$$ + v$$_{ter}$$$$\tau$$ ln (1 - x / v$$_{xo}$$$$\tau$$)

I'll skip a simplification step unless its needed: Used taylor's series which allowed cancellations.

y=v$$_{yo}$$x / v$$_{xo}$$ - x^2v$$_{ter}$$$$\tau$$ / 2v$$_{xo}$$$$^{2}$$$$\tau$$$$^{2}$$

Not sure if that's right or not... what is $$\tau$$ ? I don't know the answer... if you want I can try to work it out to see if I get the same thing...