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Free Fall

  1. Jul 24, 2007 #1
    I if a body is thrown vertically upward (from ground), and it reaches back to ground in 10 seconds.. then the initial velocity can be calculated by
    Vf = Vo + gt.. where the final velocity Vf is equal to zero and the maximun height is given by s = Vot + 1/2g(t^2)..

    Is these correct?
     
  2. jcsd
  3. Jul 24, 2007 #2

    G01

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    Yes these are these relationships are correct for this situation if your initial height is said to be 0.
     
  4. Jul 25, 2007 #3

    Integral

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    Your final velocity will not be zero. It should be equal in magnitude and in the opposite direction as your initial velocity.

    You have a trouble in the way you have written your first equation.

    [tex] V_f = V_0 + gt [/tex]

    A object obeying that equation will never slow down, it will only accelerate.

    Perhaps you meant

    [tex] V_f = V_0 - gt [/tex]
     
  5. Jul 25, 2007 #4

    mgb_phys

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    It is easier to think of it in two parts, up and then back down.
    The ball moving up has a final velocity of 0 at the top of the curve.
    The time to go up and back down will be identical.

    You can also think in terms of energy, at the top of the curve when it is stationary all the kinetic energy becomes potential energy.
     
  6. Jul 25, 2007 #5
    thank you guys for your replies
    its actually what I thought that g should be negative, and the velocity before it hits the ground is equal to the initial velocity.

    so using the formula :

    to calculate for the initial velocity is:

    -Vo = Vo - gt... and the answer is 49.05 m/s

    where t is 10 s.
     
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