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2D Kinematics questions -been stuck fo hours on it

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Okay so here is the problem: A golf ball is struck and leaves the ground at 96 m/s on a 60 angle. Determine the velocity of the ball 2 seconds after it was struck. I have the answer provided, which is 32.6 m/s at 45° above the horizon.


    2. Relevant equations

    Obviously I need to use the kinematic equations to solve this, and I have a hunch that I will need the Pythagoras as well.

    3. The attempt at a solution
    So what I have done is split the initial velocity of 96 m/s into a horizontal component and a vertical component. the horizontal being 48 m/s and the vertical 83.14. and now I am at a loss on how to proceed next.... Not sure if I am over thinking the solution, but please help
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 1, 2011 #2

    PhanthomJay

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    You can now look in the y direction and use the appropraite kimematic equation to solve for vy at t=2. Then look in the x direction to solve for vx at t=2.
     
  4. Dec 1, 2011 #3
    Sorry, I might be a little simple, but I only have the four main equations infront of me and I just cant figure out what one to use. and how do I use a kinematic equation to get a degree answer??
     
  5. Dec 1, 2011 #4
    In projectile motion:

    the horizontal component is assumed to have no acceleration and thus the velocity is constant.

    the vertical component is assumed to have only the acceleration due to gravity acting on it and thus you can use the equation vy2 = vy1 + 2at since acceleration due to gravity is acting downwards, the equation becomes vy2 = vy1 - 2gt, where vy2 is your final vertical velocity, vy1 is your initial vertical velocity, g is acceleration due to gravity(9.8) and t is time

    EDIT: after obtaining the vertical and horizontal components of velocity you must use vector addition to solve for the velocity

    the angle is the arctangent of the vertical component of velocity divided by the horizontal component of the velocity. The angle you obtain can then be expressed as [x degrees above the horizontal]
     
    Last edited: Dec 1, 2011
  6. Dec 1, 2011 #5

    gneill

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    attachment.php?attachmentid=41428&stc=1&d=1322786895.jpg
     

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  7. Dec 1, 2011 #6

    PhanthomJay

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    That's in the y direction, vy = vy_initial + ayt where ay = -g, and in the x direction, vx = vx_initial + axt where ax = 0.
     
  8. Dec 2, 2011 #7
    Okay So I took you guys advise and i did as follows:

    vy = vo - 2at where a is equal to negative gravity
    vy = 83.14 m/s - 2(-9.81m/s2)(2s)
    Vy = 122.34

    Vx = Vo - 2at where the vertical acceleration is 0
    Vx = 48 m/s -[STRIKE] 2(0)(t)[/STRIKE]
    Vx = 48 m/s

    tan ° = 48/122.34
    [STRIKE]tan-1+ tan-1[/STRIKE]°= tan-1(48/122.34)
    ° = 21.42

    but these are not the answers that I need to be ending up at. My answers have been provided for me they should be 32.6 m/s at 45° above the horizon. I have indeed tried these steps above before I posted, and gotten the same answers, I just don't know where I am going wrong
     
  9. Dec 2, 2011 #8

    PhanthomJay

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    I corrected the typos in the kinematic equations but I guess you missed it. So here they are again:

    In the y direction,
    vy = vy_initial + ayt, where ay = -g,
    and in the x direction,
    vx = vx_initial + axt, where ax = 0.
    Watch plus and minus signs.....

    where did those answers come from.....they are way off..?
     
    Last edited: Dec 2, 2011
  10. Dec 2, 2011 #9
    The answers are provided from my teacher, I was wondering as well if the answers provided where correct. I had a sneaky suspicion that they where not, as I do not normally struggle this bad on a question like this.

    Maybe I will just confirm with my teacher, before I get anymore into the question.

    than answer I keep getting to is (now that I've corrected my signs) , that the velocity is 43.94 m/s at 47.53° above the horizon?? Does this sound more correct
     
  11. Dec 2, 2011 #10

    PhanthomJay

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    I don't know how you are getting that answer...checking the maths,
    In the y direction,
    vy = vy_initial + ayt, where ay = -g,
    vy = 83.14 -9.8(2) = 63.54 m/s
    and in the x direction,
    vx = vx_initial + axt, where ax = 0,
    vx = 48 + 0 = 48 m/s
    V = sqrt(63.542 + 482)
    V = 80 m/s
    Tan-1θ = 63.54/48 = 1.32
    θ = 53°
     
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