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Projectile motion

  1. Jun 11, 2010 #1
    1. The problem statement, all variables and given/known data

    A ball is thrown at a velocity of 10 m/s, at an angle of 30° from horizontal and released 2 m
    above the ground. What will the magnitude of ball’s resultant velocity be at the moment of
    impact with the ground?

    2. Relevant equations
    Vf2 = vi2 + 2as


    3. The attempt at a solution

    Vf2 = 5(2) + 2 x -9.8 x -2
    Vf2 = 25 + 39.2
    Then find the square root

    The answer is 11.8 m/s
    What am I doing wrong, please help

    Thanks
     
  2. jcsd
  3. Jun 11, 2010 #2

    rock.freak667

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    Homework Helper

    Split the velocity into components.

    Use v2=u2-2g(s-s0) to find the vertical component of velocity when it hits the ground.
     
  4. Jun 11, 2010 #3
    Im sorry, im still stuck, could you write it out in the formulae, Im clueless when it comes to physics.
     
  5. Jun 11, 2010 #4

    rock.freak667

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    Homework Helper

    v2=u2-2g(s-s0)

    v= final velocity
    u= initial velocity
    g= acceleration due to gravity
    s= displacement
    s0=initial displacement
     
  6. Jun 11, 2010 #5
    you'll have to analyze motions in the x & y direction separately as acceleration is along the y direction only.

    X
    [itex]{v}_{x0} = 10 \cos 30[/itex]
    [itex]{a}_{x} = 0 [/itex]
    [itex]{v}_{x} = {v}_{x0} [/itex]

    Y
    [itex]{v}_{y0} = 10 \sin 30 [/itex]
    [itex]{a}_{y} = -g [/itex]
    [itex]{s}_{y} = -2 [/itex]

    [itex]{s}_{y} = {v}_{y0}t + \frac{1}{2}{a}_{y}{t}^{2} [/itex]
    solve for t & then use -
    [itex]
    {v}_{y} = {v}_{y0} + {a}_{y}t[/itex]
     
  7. Jun 11, 2010 #6
    Thanks heaps for your help =)
     
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