Titled reference frame, N2L with position and velocity

AI Thread Summary
In this discussion, participants analyze the motion of a ball thrown up an inclined plane, focusing on applying Newton's second law and deriving the ball's trajectory. They clarify that the ball's initial velocity components can be determined using trigonometric functions based on the launch angle relative to the incline. The conversation emphasizes the need to express the ball's position over time and find the intersection with the inclined plane to calculate the range. Key formulas for the range and maximum range are derived, with participants discussing the correct application of kinematic equations and the effects of gravity on the motion. Ultimately, the discussion aims to clarify the problem-solving approach for projectile motion on an incline.
  • #51
oops! little pen and ink mistake over here..

t = vosin(theta) / (1/2)gcos(phi)
yep!
 
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  • #52
Oblio said:
oops! little pen and ink mistake over here..

t = vosin(theta) / (1/2)gcos(phi)
yep!

cool. plugging that into dx should work.
 
  • #53
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)vosin(theta)/(1/2)gcos(phi)
 
  • #54
Oblio said:
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)vosin(theta)/(1/2)gcos(phi)

you missed t^2... didn't square t.
 
  • #55
oops.
man I am bad..
 
  • #56
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)(vosin(theta)/(1/2)gcos(phi))^2

on the right side i can cancel out the (1/2) on the bottom and top, as well as the g, giving,

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vosin(theta)/cos(phi))^2
 
  • #57
Oblio said:
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - (1/2)gsin(phi)(vosin(theta)/(1/2)gcos(phi))^2

on the right side i can cancel out the (1/2) on the bottom and top, as well as the g, giving,

dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vosin(theta)/cos(phi))^2

careful the 1/2 and g are squared...
 
  • #58
oops again.

so I am left with
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vo^2 sin(theta) ^2 / (1/2) g (cos(phi)^2)
 
  • #59
Oblio said:
oops again.

so I am left with
dx = vocos(theta)vosin(theta)/(1/2)gcos(phi) - sin(phi)(vo^2 sin(theta) ^2 / (1/2) g (cos(phi)^2)

Looks good... try to simplify and use a trig identity to get it to look like the formula they gave for the range...
 
  • #60
will trig identities apply since theta and phi are present?
 
  • #61
im confused how one would ever get, for example : cos (theta + phi) through simplifying
 
  • #62
Oblio said:
im confused how one would ever get, for example : cos (theta + phi) through simplifying

try a little factoring of your equation also... look up the identity for cos(A+B)...
 
  • #63
i found that cos (a+b) = cosacosb +/- sinasinb... but i don't have that relationship anywhere
 
  • #64
Oblio said:
i found that cos (a+b) = cosacosb +/- sinasinb... but i don't have that relationship anywhere

first write everything over 1 denominator... then compare what you have to the formula you need to get... a little factoring will give you the answer.
 
  • #65
k I am at
dx= vo^2sin(theta)*(cos(theta)-sin(phi)) / (1/2)cos(phi)^2(sin(theta))
 
Last edited:
  • #66
Oblio said:
k I am at
dx= vo^2sin(theta)*(cos(theta)-sin(phi)) / (1/2)cos(phi)^2(sin(theta))

factoring out the sin(theta) was correct... but you made a mistake somewhere...
 
  • #67
i can't factor out a vo^2?
 
  • #68
Oblio said:
i can't factor out a vo^2?

yes you can... I was referring to the sin's cos's... check your work... your denominator should only have [cos(phi)]^2
 
  • #69
i edited that in by mistake. i meant to put that in the numerator.

dx= vo^2sin(theta)*(cos(theta)-sin(phi)sin(theta) / (1/2)gcos(phi)^2
 
  • #70
Oblio said:
i edited that in by mistake. i meant to put that in the numerator.

dx= vo^2sin(theta)*(cos(theta)-sin(phi)sin(theta) / (1/2)gcos(phi)^2

almost there... should be cos(theta)cos(phi) in the numerator... I think you forgot to multiply the top by cos(phi) when putting everything over 1 denominator.
 
  • #71
ya, you mean i didnt multiply it by (1/2)gcos(phi)^2?
 
  • #72
ignore that last one
 
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