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Outruning an arrow?

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data

    Hi, From a previous question I have just worked out that the speed of the arrow launch is 20ms-1. The speed the mans can run is 7.0ms-1. Question, find the minimum angle above the horizontal at which the man must aim in order to outun the arrow. (you may assume the man is at rest when he fires the arrow and you may ignore the short times it takes him to accelerate from rest to his running speed).

    Please help, I have been looking at this question for about a week now and I can't see for the life of me how you can possibly work this out without knowing the distance or anything else??? I would be grateful if someone could please at least point me in the right direction. I'm doing a modular degree, 2nd year but this is my first physics course...ahhhhhh So please explain in detail, my maths is somewhat limited. I think I should have stuck to biology!!!!!!


    Many thanks,

    Claire


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 8, 2008 #2

    Andrew Mason

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    What is the horizontal component of the arrow's velocity when its velocity is 20 m/s at an angle, alpha, above the horizontal?

    AM
     
  4. Dec 8, 2008 #3
    Hi,

    Thanks Andrew, I'm still trying to digest what you said but thanks for replying.

    Would Vox = Vo Cos theta and Voy = Vo sin theta be on the right track???

    I worked out the distance the man's runs as 49m, is this right and is it really relevant. I'm not quite sure how or why I got that now, I'm starting to confuse myself now!!!
     
  5. Dec 8, 2008 #4

    Andrew Mason

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    So if Vo Cos theta were to be greater than the man's running speed, what would occur?

    AM
     
  6. Dec 8, 2008 #5
    Hi,

    I got an answer of 5 degrees as the minimum angle, which seems very small. Is this right?
    I used R = (Vo^2/g) sin 2theta

    2theta = sin-1 (9.8 x 7.0)/20 = 9.9

    divide by 2 to give theta = 5 degrees

    Many thanks,
    Claire
     
  7. Dec 8, 2008 #6
    think about what andrew mason asked you...

    if the runner only has an x-component of velocity, is the y-component of the arrow relevant?
     
  8. Dec 8, 2008 #7

    LowlyPion

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    It's really simpler than that.

    Note that if he is faster than Vo*Cosθ as Andrew pointed out then you are done.

    So if Cosθ ≤ 7/20

    or if arcCos(7/20) ≤ θ

    EDIT: Corrected egregious typo.
     
    Last edited: Dec 8, 2008
  9. Dec 8, 2008 #8
    Hi,

    Do you mean cos theta = 7.0/20 = 0.35
    giving theta = 69.5 degrees ????

    Which would be much simplier, not that I understand it any better, but here's hoping.

    Many thanks,

    Claire
     
  10. Dec 8, 2008 #9

    LowlyPion

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    Yes of course I do.

    Sorry for the typo.
     
  11. Dec 8, 2008 #10
    Hi,

    Does that mean it's right!?!? ;)

    I think I finally got it, at last!!! Thank you so much for all your help everybody

    Claire
     
  12. Dec 8, 2008 #11

    LowlyPion

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    Cheers then.
     
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