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How far will the player be if she continues to run after the ball

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A soccer player running at 6.8 m/s kicks a soccer ball straight out in front of her so that it travels with an initial velocity of 22m/s at an angle of 50° with respect to the horizontal. If the soccer player continues to run at 6.68 m/s in the same direction that the ball was kicked, how far will she be from the ball when it strikes the ground?

    2. Relevant equations

    dv=1/2*at2
    dh=Vh*Δt
    Kinetic equation d=Vi*t+ 1/2*at2
    dh=-V2*sin2θ /g
    sinθ=opp/hyp
    cosθ=adj/hyp
    Δt=-2Vsinθ / g

    3. The attempt at a solution

    First, I started by finding the length of time the ball was in the air before hitting the ground.

    dh=cos(50)*22=14.1m
    dv=sin(50)*22=16.9m

    Δt=14.1m/22m/s=0.64s

    Then I found the distance traveled by the ball
    d= (22m/s)*(0.64s)+ 1/2 (9.8)*(0.64s)2= 16.1m

    and then I calculated the distance traveled by the player
    d=(6.68m/s)*(0.64s)=4.28m I used this equation because there was no vertical distance to be measure for the player.

    Then I found that the player would be d=16.1m-4.28m= 11.8m from the ball

    I would like for someone to verify that the proper equations were used and that the significant figures are being respected. Thanks so much in advance!
     
  2. jcsd
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