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Newtons law question (Baseketball player)

  1. Nov 26, 2009 #1
    Question reads: An exceptional vertical jump from rest would raise a person 0.80m off the ground. to do this what constant force would a 70.0 kg person have to exert againts the ground. Assume the person lowers himself by 0.20m prior to jumping and remains in a standing position while in the air.

    My progress: well firstly i drew a free body diagram of him on the floor. (Fn[up] and Fg[Down]) and when he is in the air (Fg[Down])
    then i tried finding out the velocity and hopefully i can find acceleration afterwards.

    i used vf^2 = vi^2 + 2ad
    i set accelration as (-9.8m/s^2) because once he is in the air gravitational force pulls him down.

    am i going in the wrong direction ? where should i begin this problem. im also confused why crouching 0.20m prior to jumping would change anything.

    please further guide me with this problem. thanks.
     
  2. jcsd
  3. Nov 26, 2009 #2

    Integral

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    You have 2 problems to deal with, first is a freefall problem to find what velocity you need to give the required height. Then you need to find the acceleration required to give you that velocity.
     
  4. Nov 26, 2009 #3
    ok, now i understand that you can use the same equation to find the velocity and acceleration by plotting the numbers correctly.

    so i found the velocity by using equation vf^2 = vi^2 + 2ad
    vf^2 = 0 +2(-9.8)0.8
    vf = 3.95m/s
    (but question said he crouched 0.2m where does this go ??)


    using this velocity i sub it back in 3.95 = 0 + 1.6a
    a = 2.46m/s^2

    with this acceleration i use Newton's 2nd law

    Fnet = ma
    Fapp - Fg = ma
    Fapp = ma +Fg
    Fapp = 70(2.46) + 70(9.8)
    fapp = 858.2N

    But wait,The answer sheet says it should be 3400N or 340N (my teacher writing is messy)
    what did i possibly do wrong here?
     
  5. Nov 26, 2009 #4

    Integral

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    This looks good to here

    Where did this come from, please show us.

    Here is where you need to use the .2m crouch. You have a zero inital velocity and the final velocity you have found above. You need to find the acceleration which works with these values.
    with this acceleration i use Newton's 2nd law

    Fnet = ma
    Fapp - Fg = ma
    Fapp = ma +Fg
    Fapp = 70(2.46) + 70(9.8)
    fapp = 858.2N

    But wait,The answer sheet says it should be 3400N or 340N (my teacher writing is messy)
    what did i possibly do wrong here?[/QUOTE]
     
  6. Nov 26, 2009 #5
    ah ha... there was my problem.

    corrected answer: Vf^2 = vi^2 + 2ad
    3.95^2 = 0 + 2a(0.2)
    15.6025 = 0.4a
    a= 39m/s^2


    Fnet = ma
    fapp - mg = ma
    fapp = ma + mg
    fapp = 2730 + 686
    fapp = 3400N

    i made a mistake using 0.8m for the distance rather than 0.2m
    but im not sure why this is the case. he achieved the velocity of 3.95m/s during his flight.
    not just in a mere distance of 0.2m ? can you clarify this concept more clear?
     
  7. Nov 26, 2009 #6

    Integral

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    When jumping you can only accelerate upwards as long as your feet are on the floor. As soon as your feet leave the floor, gravity takes over and you now accelerate at g.

    As soon as your legs are straight your feet leave the floor. That occurs at .2m.
     
  8. Nov 26, 2009 #7
    great, thanks for your help :D
     
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