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Homework Help: Finding time and speed at which skydiver hits the ground!

  1. Mar 16, 2010 #1
    If initial vetical velocity is zero, at what time and how fast does a skydiver hit the ground? Assume that g=10m.s^-2

    Given that:
    y' = -gt + A y' is the vertical velocity
    y = -(1/2).g.t^2 +At +B y is the distance above the ground

    In order to solve how long takes skydiver to hit ground I did the following:

    0 = -(1/2).g.t^2 +(0)t + 4000
    4000 = -(1/2).g.t^2
    8000 = g.t^2
    800 = t^2
    t = 28.284 seconds!!!

    Now from here I am unsure how to find the speed at which hits the ground. Help needed sorry for bad formatting!
  2. jcsd
  3. Mar 16, 2010 #2


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    How about using v=u+at?
  4. Mar 16, 2010 #3
    Okay so if initial velocity is zero I get:

    v = 0 +10.28.284
    v = 282.84 m.s^-2
    Is this correct?

    It seems bit excessive as 282.84m.s^-2 is 632.319 MPH. DOes that seem correct to you?
  5. Mar 16, 2010 #4
    I must also say the skydivers parachute fails to open so a higher speed on descent would be expected.
  6. Mar 16, 2010 #5
    Anymore ideas please, help desperatley needed!
  7. Mar 16, 2010 #6


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    Looks ok to me if the initial height is 4000m. 282m/s IS fast, but you've completely neglected air friction. That would slow the skydiver down a lot, even if the chute didn't open.
  8. Mar 16, 2010 #7
    Ah yes thanks for the post :)
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