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Projectile Motion boat

  1. Oct 6, 2008 #1
    1. The problem statement, all variables and given/known data
    A boat is moving to the right with a constant deceleration of .3 m/s^2 when a boy standing on the deck D throws a ball with an initial velocity relative to the deck which is vertical. the ball rises to a maximum height of 8 m above the release point and the boy must step forward a distance d to catch it at the same height as the release point determine. (a) the distance d, the relative velocity of the ball with respect to the deck when the ball is caught. (b) Relative velocity of the ball with respect to the boat when caught.


    2. Relevant equations



    3. The attempt at a solution
    I do have a copy of the answers, but iv been looking over these for sometime and still dont understand a certain step taken. The text in bold/italics is the section i dont understand.

    Part (a)
    Vy = 0 when at max height of 8m

    Use
    Vy-(Vy)o = a*t
    Vy = 12.53m/s

    Finding t for whole motion
    Y-Yo = (Vy)o*t+1/2*a*t^2
    t = 2.55s

    Vx = (Vx)o+a*t (a=0.3ms^-2, deck accel)

    Distance Travelled By Deck:

    Xd = (Vx)o*t+1/2*a*t^2

    Velocity of ball with respect to deck

    Vb/d = (Vx)o-Vx

    What is the above equation based off? Relative velocity would be Vb/d = Xb/d-(Xb/d)o wouldnt it? Also the horizontal component of balls velocity is constant, so wouldnt
    Vx = (Vx)o? Making the RHS = 0? I tried using constant acceleration equations and like V-Vo=a*t and
    x-xo=Vo*t+1/2*a*t^2 to cancel out either Vx or Xd but this results in a complex number or Vx or Xd becomes canceled out.

    (Vb/d)x = (Vx)o-[(Vx)o+a*t]
    (Vb/d)x = -a*t [Eq. (1)]

    Xb/d = (Vx)o*t-Xd
    Xb/d = (Vx)o*t-[(Vx)o*t+1/2*a*t^2]
    Xb/d = -1/2*a*t^2 [Eq. (2)]

    At the time of catch Xb/d = d


    From Eq (2)
    d = -1/2*a*t^2
    d = 0.975m

    Cheers
     
  2. jcsd
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