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Body fall kinematics

  1. Nov 27, 2007 #1
    1. The problem statement, all variables and given/known data
    A body falls from a height of 409.5m under gravity. Determine:
    (a) The time taken to reach the ground, if the initial velocity is zero
    (b) The velocity at impact with ground.


    2. Relevant equations
    s= ut
    v= u+at
    s=ut + 0.5*at^2


    3. The attempt at a solution
    i tried using the formulas above but couldn't get anywhere, i guess i need to use 9.8 for the acceleration due to gravity too?
     
  2. jcsd
  3. Nov 27, 2007 #2

    stewartcs

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  4. Nov 27, 2007 #3
    i think i've just to use the equations above and mybe this one too: v^2=u^2+2as
     
  5. Nov 27, 2007 #4

    stewartcs

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    They are the same, just different notation. The link was to help clarify how to approach the problem.

    Find the time it takes for the object to hit the ground first (assuming free-fall on earth). The link will help you there.

    Then use that time to determine the velocity (part b of the question).
     
  6. Nov 27, 2007 #5

    stewartcs

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  7. Nov 27, 2007 #6

    stewartcs

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    BTW, don't forget that the problem states that the initial velocity is zero (it will reduce the equation a little).
     
  8. Nov 27, 2007 #7
    damn, i still can't figure it out, it may be the notation that's confusing me but i don't know.
    I have the info from the question: The initial velocity is 0, displacement is 409.5m(ithink)
    and the acceleration is 9.8.

    I just can't seem to use these values to come up with the time.
     
  9. Nov 27, 2007 #8

    stewartcs

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    If you look at the first link I gave you, you'll see how this is derived (which is the most important part).

    t = sqrt(2*h/g)

    where,

    t = time it takes for the object to reach the ground
    h = the height the object is dropped from
    g = gravitational accel.
     
  10. Nov 27, 2007 #9
    .5 * 9.8 * t^2 = 409.5

    t = squareroot(409.5/(.5*9.8))

    9.8 * t = answer for b and t is the answer for a
     
  11. Nov 27, 2007 #10

    stewartcs

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    The point is not to give the answer to the OP, but rather help them find it themselves so they will actually learn something.
     
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