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## Titled reference frame, N2L with position and velocity

 Quote by Oblio d = v0*t + (1/2)at^2
Yes. Now use the values we've calculated for v0 and a for this problem... for the x- direction... and the y-direction...
 dx = vocos$$\vartheta$$t + (1/2)(-g)t^2sin$$\phi$$ dy = vosin$$\vartheta$$t + (1/2)(-g)t^2cos$$\phi$$
 i have a feeling this is wrong for dy....

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 Quote by Oblio i have a feeling this is wrong for dy....
Looks right to me. So now you want to find the range...
 range?

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 Quote by Oblio range?
the question asks for the range...
 which is really just the displacement up the plane?

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 Quote by Oblio which is really just the displacement up the plane?
yes, maximum displacement... the object is thrown at an angle, then hits the incline eventually... the displacement up the plane when it hits is the range...
 is this the same kinematics equation?

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 Quote by Oblio dx = vocos$$\vartheta$$t + (1/2)(-g)t^2sin$$\phi$$ dy = vosin$$\vartheta$$t + (1/2)(-g)t^2cos$$\phi$$
use these equations to find dx when dy = 0... that gives the range.
 solve for...... vo? and insert into dx?

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 Quote by Oblio solve for...... vo? and insert into dx?
no, you want to eliminate t... so solve for t and substitute into dx.
 without simplifying yet.. dx = vocos (theta) (vosin (theta) / (-1/2)gsin(phi)) + (1/2)(-g)cos(phi)((vosin(theta)/(-1/2)gsin)^2) ya?

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 Quote by Oblio without simplifying yet.. dx = vocos (theta) (vosin (theta) / (-1/2)gsin(phi)) + (1/2)(-g)cos(phi)((vosin(theta)/(-1/2)gsin)^2) ya?
hmm... close but some mistakes... can you post what you got for t?
 t = vosin(theta) / (-1/2)gsin(phi)

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 Quote by Oblio t = vosin(theta) / (-1/2)gsin(phi)
setting dy = 0, and solving for t will give you something different... it should be cos(phi) and the - shouldn't be there.
 oops! little pen and ink mistake over here.. t = vosin(theta) / (1/2)gcos(phi) yep!