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Dynamics in Two Dimensions Question

  1. Aug 20, 2013 #1
    1. The problem statement, all variables and given/known data

    During baseball practice, you go up into the
    bleachers to retrieve a ball. You throw the
    ball back into the playing field at an angle of
    42° above the horizontal, giving it an initial
    velocity of 15 m/s. If the ball is 5.3 m above
    the level of the playing field when you throw
    it, what will be the velocity of the ball when
    it hits the ground of the playing field?

    <=42°
    vi=15m/s
    h=5.3m
    vf=?

    2. Relevant equations

    h=-0.5gt^2+VosinθΔt

    vf=vi+aΔt

    3. The attempt at a solution

    5.3m=-0.5(9.8)t^2+15(sin 42°)Δt
    t=1.02s

    vf=(15)(sin 42°)+(-9.8)(1.02)
    vf=0.041m/s

    vx^2 + vy^2 (square root)
    (15 x cos 42)^2 + 0.041^2
    =11.15m/s

    what am i doing wrong? Answer is 18m/s.
     
    Last edited: Aug 20, 2013
  2. jcsd
  3. Aug 20, 2013 #2

    lewando

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    Gold Member

    Your problem is localized here:
    The result you got for t is not correct.
     
  4. Aug 20, 2013 #3
    This looks more like projectile motion. In this case, your equation: 5.3m=-0.5(9.8)t^2+15(sin 42°)Δt is not necessary.

    It seems to be more useful as [itex]Y=Y_{o}+V_{oy}t-\frac{gt^{2}}{2}[/itex].
     
  5. Aug 21, 2013 #4

    haruspex

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    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It starts at 5.3m above the final height, so what is the change in height?
     
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