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Finding the average acceleration of a basketball player

  1. Dec 26, 2016 #1
    1. The problem statement, all variables and given/known data
    A bascketball player , who has a weight of 890 N , jumps vertically with a height of 1.2 m ,

    1- Find his speed when he left the floor.

    2- If the time that takes the basketball player to quit the floor is 0.3 s, what is his average acceleration ?

    (I don't understand this question at all, is there any function of time that I've to found or anything like this ?)

    3-Find out the value of the force that the player applies to the ground .


    2. Relevant equations
    F = m.a ..(1) , Vf²-Vi² = 2 .a.d ..(2) , p =m.g ..(3)

    3. The attempt at a solution

    1- For the 1'st question it's quite easy just by using the formula (2) above , we get
    [tex] V_{i} = \sqrt{2.g.d } = 4.84 m/s [/tex]

    2- for the second I think that it's :

    [tex] a = \frac{\Delta V}{ \Delta t} = \frac{V_{i}}{0.3} = 16.16 m.s^{-2} [/tex]

    3- For this I didn't understand at all , by using the Newton 2'nd law I get : f = m.a ( where f is the magnitude of the applied force on the ground ) , but when the player applies that force he is at rest in the ground , not moving , which means a = 0 , thus f = 0 , isn'it ???
     
  2. jcsd
  3. Dec 26, 2016 #2

    CWatters

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    Homework Helper

    Think about how someone jumps. They bend their knees then straighten them. So most of the player isn't at rest.
     
  4. Dec 26, 2016 #3
    I didn't understand , they bend their knees and straighten them , yes , and after that ? I mean how this could help us to find out f ????
     
  5. Dec 26, 2016 #4
    Maybe it would help to think of it as similar to a mass attached to the top of a compressed spring sitting on the ground (except, in the case of the spring, the force and acceleration are not constant). When the spring is instantaneously released from its compressed state, the mass starts accelerating upward, even though the bottom of the spring remains in contact with the ground. Once the spring has released all of its energy, the spring lifts up off of the ground - like the basketball players feet leave the ground. So apparently, you already calculated the acceleration, and you know the mass of the basketball player . . .
     
  6. Dec 26, 2016 #5

    CWatters

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    Your use of F=ma was correct.
     
  7. Dec 26, 2016 #6

    NascentOxygen

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    During the 0.3s take-off phase he is accelerating his Centre of Mass by pushing his feet against the floor. The moment his toes lose contact with the floor he can no longer gain acceleration—there's nothing now to be pushed against.

    https://www.physicsforums.com/attachments/110502.gif
     
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