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1-D kinematic flea

  1. Sep 15, 2011 #1
    1. The problem statement, all variables and given/known data
    If a flea can jump straight up to a height of 0.510 m, what is its initial speed as it leaves the ground? How long is it in the air?

    2. Relevant equations
    The four kinematics equations with constant acceleration I think.

    3. The attempt at a solution

    I let the positive y axis be the jump height and the x axis time.

    From there I list some known variables:
    [itex]a_{y} = -9.8 m/s^{2}[/itex]
    [itex]y_{0} = 0 m[/itex]
    [itex]\Delta y = 0.510 m[/itex]

    I'm not really sure how to get it from here. I am assuming that he reaches the apex of y=0.510 meters at 1/2 the total air time. But I can't figure out which equation to use in this case.
     
  2. jcsd
  3. Sep 15, 2011 #2
    I am pretty sure this is the equation to use here.

    [tex]v_{f}^{2} = v_{0}^{2} + 2(a)d[/tex]

    Do I assume his final velocity is at the apex of the jump, thus?

    [tex]0^{2} = v_{0}^{2}+ 2(-9.8)(0.510)[/tex]

    So his initial velocity would be:

    [tex]v_{0} = \sqrt{2(9.8)(.510)}[/tex]
     
  4. Sep 15, 2011 #3

    lewando

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    Gold Member

    Since you don't know time, which of the 4 equations does not rely on time?

    [edit] You beat me to it. Looks like you are on the right track!
     
  5. Sep 15, 2011 #4
    The one above, making initial velocity = 3.16 (to two significant figures)


    I think that opens up the possibility of using this one:

    [tex]x(t) = x_{0} + tv_{0} + \frac{at^{2}}{2}[/tex]
    ?

    So:
    [tex].510 = 0 + 3.16t + 4.9t^{2}[/tex]

    therefor t= .13368 (about 0.134 seconds)

    That seems like a small timeframe to me, which is why I am wondering if I am on the right track here.

    Edit: Yeah that is incorrect, I don't see what I am doing wrong here. Oh wait it should be -9.8, giving it another shot..
     
    Last edited: Sep 15, 2011
  6. Sep 15, 2011 #5
    Okay, that can't be it because that quadratic equation would not be solvable.
     
  7. Sep 15, 2011 #6
    Oh my, I had the right time, I just needed to multiply it by 2 to represent the other 1/2 of the jump because t was giving me the time at the top....

    Using -9.8 in the above gave me t = .3224 so really it was 6.448

    Ugh.

    Thanks
     
  8. Sep 15, 2011 #7

    lewando

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    Gold Member

    You need to be careful when evaluating -4.9t2 +3.16t -0.51 = 0. The B coefficient in not "3.16", but rather [tex]\sqrt{2*9.8*0.51}[/tex] That's why the imaginaries pop up with calculators, applets, etc.

    Keep it simple: v = v0 + at gets you there also.

    I think you meant round trip time is 0.6448 seconds
     
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