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Kinematics of a particle motion equations problem

  1. Dec 31, 2011 #1
    1. The problem statement, all variables and given/known data
    A baseball is thrown downward from a 12.5m tower with and initial speed of 4.5m/s. Determine the speed at which it hits the ground and the time of travel.
    s = 12.5m
    u = 4.5m/s
    a = g = 10m/s^2

    2. Relevant equations

    Motion equations used
    s = ut + 0.5at^2
    v^2 = u^2 +2as

    Quadratic equation
    x = (-b +/- sqrt( b^2 - 4ac))/2a

    3. The attempt at a solution

    This is what I started with, and rearranged to put into the quadratic formular.
    12.5 = 4.5t + 0.5(10)t^2
    therefore:
    5t^2 + 4.5t - 12.5 = 0

    Where:
    a = 5
    b = 4.5
    c = 12.5

    The answer I get is 1.194 seconds
    The answer in the book says exactly half that at 0.597 seconds.

    Then for part two:
    v^2 = u^2 + 2as
    seemed the right one.

    and I consistently get 16.439m/s
    yet the answer in the book says 23.9m/s

    I would be grateful for any input on this.

    Kind Regards

    Rob K
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 31, 2011 #2

    nvn

    User Avatar
    Science Advisor
    Homework Helper

    Rob K: Both of your answers currently appear to be correct, and the book appears to be wrong.

    By the way, always leave a space between a numeric value and its following unit symbol. E.g., 12.5 m, not 12.5m. See the international standard for writing units (ISO 31-0).
     
  4. Dec 31, 2011 #3
    Thank you very much indeed for verifying this for me, I am suspecting there are a few errors, though as it is the 11th edition, I thought they might have been ironed out by now.

    Happy New Year

    Rob K
     
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