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Average acceleration of a falling ball on contact with floor

  1. Dec 18, 2009 #1
    1. The problem statement, all variables and given/known data
    A ball falls on the surface from 10m height and rebounds to 2.5m.If duration of contact with floor is 0.01 sec,then average acceleration during contact is-
    g = 10m/s^2

    2. Relevant equations
    \alpha = \frac{v_2 + v_1}{t}

    3. The attempt at a solution
    Now here
    v_{1} = \sqrt{2gh_{1}

    v_{1} = \sqrt{50}


    v_{2} = \sqrt{2gh_{2}

    v_{2} = \sqrt{200}

    \alpha = \frac{\sqrt{200+50}}{0.01}

    After solving all of this,I get the answer the answer as 1581.14 m/s^2,but the answer in the book is 2100 m/s^2.Am i doing anything wrong over here?
  2. jcsd
  3. Dec 19, 2009 #2
    You'll get the answer if you do the calculation more carefully.
    An estimate, without calculator:
    sqrt(50) approx. 7
    sqrt(200) is 10sqrt(2)= 14 (approx.)
    21/0.01 = 2100

    Your mistake: sqrt(50)+sqrt(200) is NOT sqrt(50+200)
  4. Dec 19, 2009 #3
    ohh....thanks nasu
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