- #36
Oblio
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- 0
It's all necessary with no horizontal forces?
Oblio said: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.
Oblio said:algebraically or through another situation such as (t=0) (not that one I realize though)
Oblio said:nope. lol
Oblio said: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 said:there should be vcos(theta) in the denominator in the first term.
Oblio said:sorry.
had that in ink ( i promise!)
Oblio said: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.
Oblio said:y= (v[tex]_{yo}[/tex] + v[tex]_{ter}[/tex]x) / v[tex]_{xo}[/tex] + v[tex]_{ter}[/tex][tex]\tau[/tex] ln (1 - x / v[tex]_{xo}[/tex][tex]\tau[/tex])
I'll skip a simplification step unless its needed: Used taylor's series which allowed cancellations.
y=v[tex]_{yo}[/tex]x / v[tex]_{xo}[/tex] - x^2v[tex]_{ter}[/tex][tex]\tau[/tex] / 2v[tex]_{xo}[/tex][tex]^{2}[/tex][tex]\tau[/tex][tex]^{2}[/tex]