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Homework Help: Jumping Rope

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data

    A young lady is jumping rope; she always jumps so that both feet leave the ground at the same instant of time. Each of her jumps reaches a height of 14.3 cm off of the ground. (This means that she moves upward by 14.3 cm after her feet leave the ground.) The weight of the young lady is 407 N.

    for how long is she in the air each time she jumps?

    2. Relevant equations

    d=V(t) + 1/2at2

    V=d/t

    3. The attempt at a solution

    I first calculated her velocity"

    Vf2=Vi2 + 2ah

    Vf2= 0 + (2)(9.8)(.143)
    Vf = 1.67 m/s

    Next i tried calculating time:

    d=V(t) + 1/2at2

    0.143 = 1.67(t) + 1/2(9.8)t2

    i used the quadratic formula and came up with

    t=0.071 and t=-0.411

    I knew it wasnt the negative time cause time cant be negative and 0.071 just seemed too short of a time.

    Can anyone confirm that this is correct?

    I know the answers are NOT 0.171 or 0.086 from previous attempts.

    Thank you
     
    Last edited: Feb 6, 2010
  2. jcsd
  3. Feb 6, 2010 #2

    kuruman

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    This equation should have a negative sign in front of the acceleration term because the initial velocity and the acceleration are in opposite directions.
     
    Last edited: Feb 6, 2010
  4. Feb 6, 2010 #3
    I tried that but it doesnt work because you cannot take the square root of a negative.

    -4.9t2+ 1.67t - 0.143 = 0

    quadratic formula

    1.67 +- sqrt(1.672 - 4(-4.9)(-0.143) / 2(-4.9)

    1.67 +- sqrt (2.7889 - 2.8028)

    this is impossible because the sqrt(-0.0139) is an imaginary number......
     
    Last edited: Feb 6, 2010
  5. Feb 6, 2010 #4

    kuruman

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    First off the quadratic formula is supposed to have a denominator that you do not show. Of course this does not address the problem of the negative square root. Your problem is that you have a method that does not work very well if you do intermediate calculations that introduce round off errors. If you do it right, you will end up taking the square root of zero which is zero. To do it right (following your method - there are other methods too) use symbols and plug in at the very end. Then you will see what happens.

    Following your method, the initial velocity is found from

    [tex]v^2=0^2-2(-g)h \ \rightarrow v = \sqrt{2gh}[/tex]

    Now go to the kinematic equation and find the time it takes the girl to reach maximum height h. Again following your method with the correct sign in front of g,

    [tex]h=\sqrt{2gh}\;t-\frac{1}{2}g\:t^2[/tex]

    Solve this quadratic without plugging in numbers, see what you get, then plug in.
     
  6. Feb 6, 2010 #5
    i am a little confused because if the correct sign in front of g is negative then wouldn't the

    [itex]
    \sqrt{2(-g)h)}
    [/itex]

    not work because there again is a negative sign under the square root?
     
  7. Feb 6, 2010 #6
    so even if i do use -g for the -1/2gt^2 that will make it +1/2gt^2 which i tried before and came up with 0.071
     
  8. Feb 6, 2010 #7

    kuruman

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    No. The symbol g stands for +9.8 m/s2. It represents the magnitude of the acceleration of gravity. The acceleration itself is is a = -g or - 9.8 2m/s, the minus sign indicating the direction of the acceleration which is "down". Thus the kinematic equation which is in general

    y = v0t + (1/2) a t2

    becomes in this case with a = -g

    y = v0t - (1/2) g t2

    so that when you plug in numbers, you must put +9.8 m/s2 because you already put in the minus sign to indicate that the direction of the acceleration is down.
     
  9. Feb 6, 2010 #8
    thank you for explaining that.

    if i keep the numbers in my calculator then it works out to be this:

    -1.67 +- sqrt(0) / 2(-4.9)

    so then

    t=0.1704
    t=-0.1704

    except for earlier i tried 0.171 and that is incorrect?
     
  10. Feb 6, 2010 #9

    kuruman

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    Forget the calculator. Do it symbolically as I suggested in posting #4. Then you will see what is going on.
     
  11. Feb 6, 2010 #10
    ok i did it symbolically

    i got the quadratic formula down to this:

    -sqrt(2gh) +- sqrt(0) / -g


    so this is just equal to the velocity divided by the acceleration due to gravity correct?
     
  12. Feb 6, 2010 #11

    kuruman

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    Yes. You could have gotten the same result if you used the general equation

    v = v0 - gt

    to find how long it takes the velocity to go from v0 to zero. Of course, the time the girl stays in the air is twice that.

    You could also find directly how long she stays in the air by noting that just before she lands, her velocity is -v0. Then the total time t she stays in the air is found from

    -v0 = v0 - gt

    Using the velocity equation eliminates the quadratic and its inherent pitfalls.
     
  13. Feb 6, 2010 #12
    AHHHHHH!!!!!

    "Of course, the time the girl stays in the air is twice that."

    That is what i was not taking into consideration.

    Thank you :)
     
  14. Feb 6, 2010 #13
    Could i get some help on a couple of these concept questions please that deal with this same question....

    Each time that she is on the ground, she is first landing and then jumping (pushing off). At one instant of time while she is on the ground, she is momentarily at rest (when she is in the lowest point of her crouch); her jump begins from this instant when she is momentarily at rest. If the time of the part of her jump before her feet leave the floor is always 0.260 s, what is the magnitude of her average acceleration while she is pushing against the floor?

    so i found the acceleration
    a = v/t
    a=6.42m/s^2

    so the part i need help with is this question:

    What is the direction of the acceleration just previously found?

    The answer choices are:

    upwards
    downwards.

    I picked downwards because she is pushing against the ground therefore her acceleration is downwards.

    But i am rethinking it and i think it may be upwards because she is jumping up so therefore her acceleration is upwards????





    Assuming that the young lady's acceleration is constant while she is touching the ground, then calculate the size of the force she is applying to the ground while she is touching the ground.

    I found this to be 674 N

    the question i need help with is:

    The direction of the force described in the previous part is______

    The answer choices are:

    the force is zero so no direction
    upwards
    downwards

    i picked the force is zero so no direction because this downward force of 674 N is equal and opposite to the upward force she has of 674 N.





    What is the force which is responsible for launching the young lady into the air?

    The answer choices are:

    the weight of the young lady
    the contact force on the ground by the young lady
    the force of gravity on the Earth by the young lady
    the contact force on the young lady by the ground

    i think it is the contact force on the young lady by the ground because the ground is what applies the force against her force which therefore launches her in the air....


    Let me know if i am right or wrong please!
    Thanks for any help or explanations :)
     
  15. Feb 7, 2010 #14
    Can anyone please check over these answers in the post above and let me know if they are correct. I really wanna make sure i have these right and my assignment is due tomorrow!!

    Thank you. I appreciate any help :)
     
  16. Feb 7, 2010 #15

    kuruman

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    In what direction is her velocity changing? That is the direction of her acceleration.


    A force is a vector and, unless it is zero, it always has a direction. If the girl jumped on your chest instead of on the floor in what direction would your chest move? That is the direction of the force that she exerts.

    If the floor were not there to exert a force on her, she would keep on falling, so yes, you are correct here.
     
  17. Feb 7, 2010 #16
    Thank you. I understand the last two completely now i just am still a little unsure about the first one.

    I believe her velocity is changing to the upward direction therefore her acceleration would be upwards?

    Does that sound correct?

    Thank you for the help :)
     
  18. Feb 7, 2010 #17

    kuruman

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    Yes, the acceleration is upwards. Her initial velocity is down and her final velocity is up. The change in velocity is what you need to add to the initial velocity to get the final velocity.
     
  19. Feb 7, 2010 #18
    Thank you very much :)
     
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