Titled reference frame, N2L with position and velocity

  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. I know that N2L for the ball can't just be ma=-mg (or a= -g) since it's still travelling up...
     
  4. HallsofIvy

    HallsofIvy 40,214
    Staff Emeritus
    Science Advisor

    ?? 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. Mentz114

    Mentz114 4,110
    Gold Member

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

    learningphysics 4,124
    Homework Helper

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

    learningphysics 4,124
    Homework Helper

    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. Where does that come from?
     
  12. learningphysics

    learningphysics 4,124
    Homework Helper

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

    a = -g.
     
  13. learningphysics

    learningphysics 4,124
    Homework Helper

    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. 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. learningphysics

    learningphysics 4,124
    Homework Helper

    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. 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. learningphysics

    learningphysics 4,124
    Homework Helper

    See my previous post about rotating the axes, so that the incline is horizontal...
     
  18. yeah that is what i had done originally.
     
  19. learningphysics

    learningphysics 4,124
    Homework Helper

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

    learningphysics 4,124
    Homework Helper

    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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