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Basketball player jump physics problem

  1. Oct 13, 2005 #1
    Would somebody be able to go over my answers to this question. There are a couple of them that I am not sure of. Thank you!!
    A 100kg baseketball player can leap straight up in the air to a height of 80cm. You can undersdand how by analyzing the situation as follows:
    a. The player bends his legs until the upper part of his body has dropped by 60cm, then he begins his jump. Draw separate free body diagrams for the player and for the floor as he is jumping but before his feet leave the ground.

    There are only two forces when his feet are above to leave the ground, Fn and Fg.
    The force pointing down is his weight plus ma and using Newton's third law, an equal but opposite reaction is pushing up on him
    b. Is there a net force on the player as he jumps (before his feet leave the ground)? How can that be? Explain.
    I'm not totally sure on this one:
    Before he leaves the ground there isn't a net force because he hasn't actually moved. His feet are still in the same place on the ground. An net force pointing the positive y direction will appear when his feet leave the ground, which pulls him up.
    c. WIth what speed must the player leave the ground to reach a height of 80cm?
    [tex] v_{2}^2 = v_{1}^2 + 2ad [/tex] where d = 0.8 m
    Therefore [tex]v_{2}^2 = 3.96 m/s[/tex]
    d. What was his acceleration, assumed to be constant as he jumped?
    [tex] 3.96^2 = v_{1}^2 + 2a(0.6) [/tex]
    therefore a = 13.068 m/s^2
    e. Suppose the player jumps while standing on a bathroom scale that reads in newtons. What does the scale read before he jumps, as he is jumping and after his feet leave the ground?
    Before his jump his weight is shown 100 kg. In Newtons it's 980N
    As he is jumping F = ma = 1306N (I'm not sure about this one...Does the acceleration of 13.068 m/s^2 already take into account the affect of g on the player? or do we have to add the affect of gravity as well?)
    After his feet leave the ground the scale reads 0N
  2. jcsd
  3. Oct 13, 2005 #2

    Doc Al

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    Staff: Mentor

    Sounds good.
    Careful. Newton's third law says that whatever force the floor exerts on the player (which is Fn), the player must exert the same force (but opposite) on the floor.
    Just because his feet don't move doesn't mean there is no acceleration of his center of mass. Compare this with your answer to part (d).
    Assuming (as you are expected to) that the player's center of mass rises 0.6 m, then this is OK. (Reconsider your answer to part b.)
    Yes, you have to take gravity into account. Newton's 2nd law says that the net force equals ma. As you state above, there are two forces acting on the player--the scale will read the normal force.
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