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Find the equation for horizontal motion

  1. Jun 19, 2008 #1
    a ball is thrown with v(initial)=5i+19.6j

    find the equation of the horizontal motion



    2. Relevant equations



    the problems asks for ymax and tymax and part b asks for the equations of the horizontal and vertical motions.

    Lost on the equations
     
  2. jcsd
  3. Jun 19, 2008 #2

    Kurdt

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    What have you tried so far?
     
  4. Jun 19, 2008 #3
    your velocity is a vector... break it into components and you will see that your horizontal vector is unrestrained unless you are accounting for air resistance and then your vertical vector will be affected by gravity based on time...
     
  5. Jun 19, 2008 #4
    No air resistance. The x component is 5 and y is 19.6. I don't know how to write the equation of the line. would it just be x=5.0? It doesn't seem that easy considering entire motion is parbolic? Would y=19.6 be correct for the motion in the y direction?
     
  6. Jun 19, 2008 #5

    Kurdt

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    Have you heard of the kinematic equations for constant acceleration? You will need to use those.
     
  7. Jun 19, 2008 #6
    no what are they?
     
  8. Jun 19, 2008 #7
    do you mean:
    xf=xi+vit+1/2at^2
    and vf=vi+at
     
  9. Jun 19, 2008 #8

    Kurdt

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  10. Jun 19, 2008 #9
    i believe you meant constant velocity... the velocity in the x direction is constant so you integrate with respect to time to get the position in terms of time... and then you have your velocity in the vertical direction which isnt constant because it is getting smaller due to gravity so you have to put that in and find WHEN the vertical velocity is zero so that you can integrate the entire vertical velocity equation to get position based in terms of time and plug in that time you just obtained to get max height...
     
  11. Jun 19, 2008 #10

    Kurdt

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    You can use the constant acceleration equations with acceleration set at 0 and they reduce to those for constant speed.
     
  12. Jun 19, 2008 #11
    yes but if you start with 0 as your acceleration and integrate to get velocity what do you get?
     
  13. Jun 19, 2008 #12
    xf=xi+vit+at2

    xi=0
    vi=5
    a=?
    t=4sec (or 2*tymax)
    Xf=0+(5*4)+8a

    yes no...I still don't know where this gets me toward writing an equation for the motion.
     
  14. Jun 19, 2008 #13
    In review of the post it appears that I didn't know the RELEVANT EQUATIONS. I did. I have found ymax to be 19.6 and tymax to be 2.0
     
  15. Jun 19, 2008 #14

    Kurdt

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    That is the equation of motion in the x direction. Acceleration is 0 by the way if you're ignoring air resistance.
     
  16. Jun 19, 2008 #15
    if you have to use those equations in part a of the question why would my teacher ask for the equation of the motion in the x and y directions in part b?
     
  17. Jun 19, 2008 #16
    post what equations you have actually arrived at and maybe we can check what you got
     
  18. Jun 19, 2008 #17
    Ymax=Yi+ViyT-4.9t^2
    Ymax=19.6t-4.9t^2

    Vf=Vi+at @ymax Vf=0 a=9.8(down)
    t=2.0=Tymax

    Ymax=0+19.6(2)-4.9(4)
    Ymax=19.6

    Correct
     
  19. Jun 19, 2008 #18
    looks good to me, thats what i got too... are you looking for anything else?
     
  20. Jun 19, 2008 #19
    he asks for the equation of vertical motion y(t) and horizontal motion x(t)

    I don't know how to express that
     
  21. Jun 19, 2008 #20

    Kurdt

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    Those are the equations of motion. I don't know why your teacher has asked for them but there we go. Its as simple as that.
     
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