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A dart is thrown

  1. Sep 17, 2016 #1
    1. The problem statement, all variables and given/known data
    A dart is thrown horizontally toward the bull's-eye, point P on the dart board in the figure, with an initial speed of 8.9 m/s. It hits at point Q on the rim, vertically below P 0.20 s later.
    a) What is the distance PQ?
    b) how far away from the dart board did the dart thrower stand?
    2. Relevant equations
    x=xi + .5(vi+vf)t
    vf^2= vi^2 + 2a(xf-xi)

    3. The attempt at a solution
    I am not really sure how to begin this problem. I know the knowns are
    vi = 8.9 m/s
    t= 0.20 s
    vf = 0 m/s
    And I believe I need to use one of the kinematic equations.
     
  2. jcsd
  3. Sep 17, 2016 #2
    Can we start with the basics. What do those Relevant Equations represent, and where should they be used ?

    To help a bit further, ##v_f## is not zero. We want velocity when the dart arrives, not when it's embedded in the board.
     
    Last edited: Sep 17, 2016
  4. Sep 17, 2016 #3
    I got the distance away from the board solved but I do not think that final velocity is needed to find the y-component from P to Q. Instead, I tried to solve for the variables related to y and got.
    vi = 0m/s
    a = -9.8 m/s^2
    t = 0.20s
    Δy = ?
     
  5. Sep 17, 2016 #4
    Okay, so what equation should be used, that you can plug those numbers into to get your Δy.
     
  6. Sep 17, 2016 #5
    you were right about the velocity. I found it with respect to y using v = vi +at and then substituting into Δy = .5(vi+vf)t which gave me the answer.
     
  7. Sep 17, 2016 #6
    k, as long as you didn't use the same ##v_i## in 'a' as 'b'. You could have gone straight to ##d=v_it+at^2/2##
     
    Last edited: Sep 17, 2016
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