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How long was it in the air before returning to Earth?

  1. Sep 22, 2006 #1
    A man jumps to a vertical height of 2.5 m. How long was it in the air before returning to Earth?

    here is what i got so far...
    vi=0
    v=0
    x-xi=2.5
    a=-9.8

    that is for the man jumping upwards...I beleive the up and down will be proportional so if I just double my answer in the end i should be alright...

    is used the equation x-xi=vi*t + .5((a)t^2)
    but I cant seem to get the right answer...
    I am thinking that vi is not 0 but instead something else!
    Any help would be great!
     
  2. jcsd
  3. Sep 22, 2006 #2

    LeonhardEuler

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    Are you sure [itex]v_i=0[/itex]? The man would not move upward if his initial velocity were 0, right? You are using the right equation, though. There is one more tricky part, but I'll let you try to figure it out before I tell you how to do it.
     
  4. Sep 22, 2006 #3
    Thanks for replying I figured vi was not 0 but the question does not reveal the true vi to the reader so vi = 7??
     
  5. Sep 22, 2006 #4

    LeonhardEuler

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    That's the other tricky part. Here's another hint. You know the man's position at two points in time. This gives you two equations.

    There is a simpler way of doing it, though. You are right that the upward and downward journeys take the same time. The initial velocity is not 0 for the upward journey, but what about the downward one?
     
  6. Sep 22, 2006 #5
    going down the init velocity should be 0 because the object is at rest
     
  7. Sep 22, 2006 #6

    LeonhardEuler

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    Right, so if you apply the equation there you won't have to worry about vi.
     
  8. Sep 22, 2006 #7
    So 2.5 = -19.6t^2 or -.128 =t^2
    so time = .35 or something? is that correct?

    and is displacement negative on the way down?
    The answer has to be somewhere around 1.4...am i hot or cold?
     
    Last edited: Sep 22, 2006
  9. Sep 22, 2006 #8

    LeonhardEuler

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    If your taking the upward direction as positive, then x-xi=-2.5, not 2.5 because the man starts higher than he ends up on durring the downward trip. Also, its .5a*t^2, not 2a*t^2. And finally, remember that this is only half the journey, so you'll have to double whatever answer you get.
     
  10. Sep 22, 2006 #9
    Yea im getting about 1.43 which is what I originally got perhaps my webassign is just messed up and wont accept a good answer thanks for the help!!! :biggrin:
     
  11. Sep 22, 2006 #10

    LeonhardEuler

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    Yes, that's what I get, too.
     
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