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Hang time

  1. Sep 11, 2007 #1
    An athlete jumps vertically. treat athlete as particle and Ymax is the maximum height above the floor the athlete achieves. To explain why he seems to hang in the air, calculate the ratio of the time he is above ymax/2 to the time it takes him to go from the floor to that height.

    we get 2 equations,
    Vymax/2 = V0Y-gt_1
    Vymax = Vymax/2 - gt_2

    after isolating t
    we get:
    2t_2 / t_1 = 2 Vymax/2 / V0y-Vymax/2

    V0Y - Vymax/2 < Vymax/2
    <<<< how do you go about explaining this in common language??

    the time it takes for person to reach ymax/2 from Yground is les thatn time it takes the person to reach Ymax from Ymax/2, so Vavg from Yground to Ymax/2 is greater thatn Ymax/2 to Ymax.
  2. jcsd
  3. Sep 11, 2007 #2


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    You already know that Vymax = 0

    Use gravitational potential energy to calculate the velocity at ymax/2 in terms of v0

    Then you'll know the velocity at the bottom... v0, velocity at ymax/2, and the velocity at ymax which equals 0.

    so then you can calculate the times... why did you 2t_2 / t_1 instead of t_2/t_1?
  4. Sep 11, 2007 #3
    i set t_2 as time from ymax/2 to ymax. so the time above ymax/2 is 2t2.
    gravitational potential??
    mgh = mv^2/2??
  5. Sep 11, 2007 #4
    on this problem, i'm only allowed to use kinematics. as much as i want to use energy, i can't.
    Last edited: Sep 11, 2007
  6. Sep 11, 2007 #5


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    EDIT: Never mind you're absolutely right about the 2t_2/t_1. I misunderstood the question.

    Yeah, mgh = mv^2/2 was what I was thinking.

    You can also get the velocity at ymax/2 using...

    vf^2 = vi^2 + 2as
    Last edited: Sep 11, 2007
  7. Sep 11, 2007 #6
    ya but t1 is time from bottom to middle. the question asks for 2t2/t1 not t2/t1. but anyways, thats technical... no big deal.

    wouldn't using vf^2 = vi^2 +2as make this more complicated than it is??

    i already got a ratio of 2Vymax/2 to Voy-Vymax/2
    there is a way to simplify that?? by substituting in another equation. I dont see how it would simplify it. maybe i'm just bad at algebra, can you show me?
  8. Sep 11, 2007 #7


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    Yeah, you're right. I was misunderstanding.

    2t_2 / t_1 = 2 Vymax/2 / V0y-Vymax/2

    This doesn't simplify unless you write Vymax/2 in terms of V0y.... the question wants you to get the actual number for this ratio...

    Also be careful not to confuse Vymax/2 with (Vymax)/2 = 0 (since Vymax = 0)... might be better to write Vx where x = ymax/2... or something like that.
  9. Sep 11, 2007 #8
    vx^2=v0y^2 + 2 a ymax/2

    you get:
    2 sqrt (Voy^2-2gYmax/2) to Voy- sqrt (V0y^2-2gymax/2) RATIO

    how do you simplify that??
  10. Sep 11, 2007 #9


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    Use the same equation again, but this time to get ymax in terms of v0. ie: vx = 0, s = ymax.
  11. Sep 11, 2007 #10
    Ymax = Voy^2/2g.
    sub in and i get
    2sqrt(2gYmax/2) to V0y-sqrt(2gYmax/2)

    sub in V0y

    and i get
    2sqrt(2gYmax/2) to sqrt (2gYmax) - sqrt (2gYmax/2)

    some algebra help... how you simplify whats in bold...

    if i'm doing it right.... i get 2 to 1 ratio?? that doesn't sound right.
  12. Sep 11, 2007 #11


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    sqrt(gYmax) cancels from the numerator and denominator...

    You should get: [tex]\frac{2}{\sqrt{2} - 1}[/tex]
  13. Sep 11, 2007 #12
    wait... i did it again and got ratio of
    2 to sqrt2 - 1
    is that right??
  14. Sep 11, 2007 #13
    i guess i did it right..
    i knew the physics of this problem. but i couldn't figure out the algebra.
    greatly appreciate your algebra lesson. thank you very much.
  15. Sep 11, 2007 #14


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    no prob! good job!
  16. Sep 11, 2007 #15


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    yeah, that's what I got.
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