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Titled reference frame, N2L with position and velocity

  1. Sep 14, 2007 #1
    A ball is thrown with initial speed vo up an inclined plane. The plane is inclined at an angle(fi) above the horizontal, and the ball's initial vecity is at an angle (theta) above the plane. Choose the axes with x measured up the slope, y normal to the slope and z across it. Write down Newton's second law using these axes and find the ball's position as a function of time. Show that the ball lands a distance

    R=2vo^2(sin(theta)cos(theta+fi))/gcos^2(fi)

    from its launch point. Show that for given vo adn (fi), the maximum possible range up the inclined plane is

    Rmax=vo^2/g(1+sin(fi))

    [Don't know if I'm spelling fi right, I mean this symbol : [tex]\phi[/tex]



    How do I approach this? I think the question is written poorly, they mean that the ball is thrown in the air and it lands ON the plane.
    Without knowing the velocity how do I find the parabolic motion the ball will make?
     
  2. jcsd
  3. Sep 14, 2007 #2
    I know that N2L for the ball can't just be ma=-mg (or a= -g) since it's still travelling up...
     
  4. Sep 14, 2007 #3

    HallsofIvy

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    ?? You know the balls inital velocity! It is v0. Further, since you are also told that the ball is initially thrown at angle [itex]\theta[/itex] above the plane you know that the balls initial x component of velocity is [itex]v_0 cos(\theta)[/itex] and the y component is [itex]v_0 sin(\theta)[/itex]. You should be able to show that the balls position at time t is [itex]x= v_0 cos(\theta) t[/itex] and [itex]y= -g/2 t^2+ v_0 sin(/theta)t[/itex] (taking (0,0) as initial position). You can solve the first equation for t: [itex]t= x/(v_0 cos(\theta)[/itex] and put that into the equation for y: [itex]y= -g/(2v_0 cos(\theta)x^2+ (sin(\theta)/cos(\theta))x[/itex] to get the equation of the parabola. Now it is a question of determing where that parabola intersects the line giving the inclined plane. That is, of course, [itex] y= tan(\phi)x= (sin(\phi)/cos(\phi))x[/itex].
     
  5. Sep 14, 2007 #4

    Mentz114

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    [tex]\phi[/tex] is spelt 'Phi'
     
  6. Sep 14, 2007 #5
    Can you explain to me how you got [itex]y= -g/2 t^2+ v_0 sin(/theta)t[/itex] ?
     
  7. Sep 14, 2007 #6
    I suppose it's mostly the /2t^2 I'm failing to see...
     
  8. Sep 14, 2007 #7

    learningphysics

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    What are the forces along the plane... what are the forces perpendicular to the plane...
     
  9. Sep 14, 2007 #8
    i see that gravity is affecting v(y), but why /2t^2?
     
  10. Sep 14, 2007 #9

    learningphysics

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    that's just for the (1/2)gt^2.

    Hint: Rotate your sketch... so that the incline is horizontal... now you can treat it like a regular projectile problem... the only differences are the vertical and horizontal accelerations...
     
  11. Sep 14, 2007 #10
    Where does that come from?
     
  12. Sep 14, 2007 #11

    learningphysics

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    From the kinematics equation d = v0*t + (1/2)at^2

    a = -g.
     
  13. Sep 14, 2007 #12

    learningphysics

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    When the projectile is in the air... the only force acting on it is gravity... what is the component of gravity along the plane... what is the component of gravity perpendicular to the plane...

    so what's the component of acceleration along the plane... what's the component of acceleration perpendicular to the plane...
     
  14. Sep 14, 2007 #13
    Sorry, duh. 'Brainfart' there Learningphysics.
    I more or less understand the steps, but could you explain the logic behind why you put the equation for x into y?

    And why is '[itex] y= tan(\phi)x= (sin(\phi)/cos(\phi))x[/itex] ' where the parabola interesets?
     
  15. Sep 14, 2007 #14

    learningphysics

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    That's HallofIvy's post, not mine. The thing is Halls is considering x along the horizontal... y along the vertical... I think the question wants you to approach the problem a little differently...

    Work the problem taking x to be along the plane... y to be perpendicular to the plane... I think it makes the problem a lot easier...
     
  16. Sep 14, 2007 #15
    That it what i was trying to do....

    so don't do it the way halls of ivy said? im not sure where to go with this now...
     
  17. Sep 14, 2007 #16

    learningphysics

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    See my previous post about rotating the axes, so that the incline is horizontal...
     
  18. Sep 14, 2007 #17
    yeah that is what i had done originally.
     
  19. Sep 14, 2007 #18

    learningphysics

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    cool. can you show how far you got?
     
  20. Sep 14, 2007 #19
    not very. Not using HallsofIvy's method I'm not sure how to figure the distance and parabolic motion of the ball.
     
  21. Sep 14, 2007 #20

    learningphysics

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    Given v0 and that the angle at which the ball is launched above the incline is theta... what is the initial velocity along the incline... what is the initial velocity perpendicular to the incline ?
     
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