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Magnitude of average acceleration

  1. Sep 1, 2010 #1
    1. A golf ball released from a height of 1.50 m above a concrete floor, bounces back to a height of 0.62 m. If the ball is in contact with the floor for 5.39 ms, what is the magnitude of the average acceleration a of the ball while it is in contact with the floor?



    2. v=v(initial)+acceleration*time
    acceleration = -9.8 m/s^2



    ]3. I don't know where to start =|. Please assist
     
  2. jcsd
  3. Sep 1, 2010 #2

    diazona

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    OK, well to start you off: the problem asks for an acceleration. What equations do you know that might allow you to find acceleration?
     
  4. Sep 2, 2010 #3
    a= (v final - v initial) / (t final - t initial)
    a = dv/dt
    a d^2(s)/d(t)^2
     
  5. Sep 2, 2010 #4
    Problem solved. 1.65*10E3 m/s^s.
    For all who may come along seeking help: initial velocity is the velocity when the ball first contacts the floor and final velocity is the velocity at which the ball leaves. Delta T is the time in contact with the floor.

    With the distanc equation you can solve for the time and then multiply by acceleration to find velocity.
    0=initial position-1/2 a t^2 solve for t
    plug t into v=at
    this is initial velocity -5.42 m.s

    final velocity ises a different equation
    v final ^2 = v(initial)^2-2a(final p- initial p)
     
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