1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    rock.freak667

    User Avatar
    Homework Helper

    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

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    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 :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook