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Kinetics question in 2 dimensions

  1. Oct 9, 2009 #1
    1. The problem statement, all variables and given/known data

    A ball is hit by a golfer. The angle is 28 degrees, the velocity of the projectile is 16.0, the distance is 16.8m on horizontal. What is the velocity before impact on the ground of the ball?

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 9, 2009 #2
    Do you expect us to help you? What is giving you trouble?
     
  4. Oct 9, 2009 #3

    tiny-tim

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    Welcome to PF!

    Hi carolyn89012! Welcome to PF! :wink:

    Use the standard constant acceleration equations in the x and y directions separately …

    what do you get? :smile:
     
  5. Oct 10, 2009 #4
    Ok, so we have:

    x=V_ox*t + 1/2 a_x t^2
    16.8=14.18* t + 1/2 (0) t^2
    16.8/14.18=t
    t=1.14

    However, a need the answer for y to continue, is that correct?

    Do I solve for y distance with tan of y/x, since I know the angle?
     
  6. Oct 10, 2009 #5

    tiny-tim

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    Hi carolyn89012! :smile:

    (try using the X2 and X2 tags just above the Reply box :wink:)
    Yes, that's right (btw, it's ok to just say "x = Voxt", without mentioning a, if ax = 0). :smile:

    hmm … I honestly don't know what you do now …

    the question doesn't seem to make sense as it is …

    you've been given too much information, and it's contradictory. :frown:

    Is that the complete question (eg, is the hole higher up than the golfer)?
     
  7. Oct 10, 2009 #6
    Y will be 0 since the ball is colliding with the ground.
     
  8. Oct 11, 2009 #7
    "hmm … I honestly don't know what you do now …

    the question doesn't seem to make sense as it is …

    you've been given too much information, and it's contradictory. "


    Yes, I agree thank you! There appears to be too much information. a_x=0, y=0, and t=1.14s after plugging in the numbers.

    I, then, solved for V_x and V_y and then combined the two in solving for V_f using the the square root of the addition of (V_x)2 + (V_y)2. But don't know if any of this is correct.
     
  9. Oct 11, 2009 #8

    tiny-tim

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    Yeah, it was going to be a stupid question anyway …

    golf balls have dimples which give them aerodynamic lift, so they don't follow the usual parabolic path. :rolleyes:

    This question doesn't seem to have an answer, so just turn in anything. :smile:
     
  10. Oct 11, 2009 #9
    Ok, thanks for the help anyway. :)
     
  11. Oct 11, 2009 #10
    Note that

    [tex] V^{2}_{x} + V^{2}_{y} = V^{2}\sin^{2}(\theta) + V^{2}\cos^{2}(\theta) = V^{2} (\sin^{2}(\theta) + \cos^{2}(\theta) ) = V^{2}[/tex]

    Using the trig identity sin^2(x) + cos^2(x) = 1
     
  12. Oct 11, 2009 #11
    Interesting point!

    A classmate and I were just trying to figure out how to calculate sin^2 or cos^2 on our calculators--couldn't figure out what to do, nor could our TA.
     
  13. Oct 11, 2009 #12
    ok first get the time..

    Δd= vot-1/2at2

    16.8= 14.127t ( the acceration for the x coordinate is 0)
    t= 1.189

    now use a= vo+vf /2 <<<<<solve for the Vfy

    -9.8= (7.51+x)/1.189


    Vfy = -19.1622m/s

    now u have to get the resultant

    √(-19.16222+14.1272)

    the answer is then -23.80m/s and at an angle of 53.601

    hope this helps...
     
  14. Oct 11, 2009 #13

    lewando

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    Am I missing something? This looks like a simple problem with a parabolic trajectory. There's been no talk of wind resistance, dimples or any x-axis acceleration in the problem statement. Therefore: wouldn't the landing be symmetrical to the takeoff? Why wouldn't the velocity on landing simply be 16 m/s at -28 degrees?
     
  15. Oct 11, 2009 #14

    lewando

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    Oops, yes I was missing something. I forgot th 18m horizontal. I'm done for the day.
     
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