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Ball in free-fall

  1. Apr 3, 2009 #1
    1. The problem statement, all variables and given/known data

    A ball of mass m = 8 kg is dropped from rest at a height h = 13.9 m above the ground. Ignore air resistance.

    (a)If the ball is being released with a downward speed 4.9 m/s initially, what will be its final speed when it hits the table 0.7 m below the release point?

    (b) If the ball is being thrown upwards instead with the same speed 4.9 m/s, what is the final speed when it hits the table?
    2. Relevant equations

    Kf + Ugf = Ki + Ugi

    3. The attempt at a solution
    vf = sqrt(v(initial)^2 + 2g(h - y))
    I just plugged in numbers and got 16.8 for part (a) and I plugged in numbers for part (b) and got 17.3. Also, for part (b), I substituted y=0 because I chose the y-axis to start at the bottom where the hand would be.
     
  2. jcsd
  3. Apr 3, 2009 #2

    LowlyPion

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    To be clear. The ball is released at 13.9m and it hits the table at .7m?

    And the initial velocity is 4.9m/s.

    What does dropped from rest mean when you have an initial velocity given?
     
  4. Apr 3, 2009 #3

    rl.bhat

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    vf = sqrt(v(initial)^2 + 2g(h - y))
    In the first part the distance traveled by the ball is only 0.7m.
     
  5. Apr 3, 2009 #4
    Ignore that released from rest part.
     
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