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Acceleration caused by bending the knees

  1. Jan 25, 2012 #1
    1. The problem statement, all variables and given/known data
    A 97.4 kg basketball player can leap straight up in the air to a height of 71.9 cm. The player bends his legs until the upper part of his body is dropped by 61.2 cm, then he begins his jump. With what speed must the player leave the ground to reach a height of 71.9 cm?
    What was his acceleration, assumed to be constant, as he jumped (before his feet left the ground)?

    2. Relevant equations
    F= ma
    player's weight = m x 9.8 m/s2= 955 N

    3. The attempt at a solution
    I got the first question (with what speed much the player leave the ground) right, vf=3.75 m/s.
    I know that the acceleration must increase to overcome the force of your weight as he jumps, but I'm unsure how to calculate what that constant acceleration would be, and how to calculate that from the values given.
    HELP PLEASE!
     
  2. jcsd
  3. Jan 25, 2012 #2
    Have you come across the constant acceleration kinematics equations?
     
  4. Jan 25, 2012 #3
    I tried the equation v2=v02+2ad.
    I'm unsure what to plug in as the distance to be covered. Initial velocity should be 3.75 as found previously, and final velocity zero (at the maximum height).
    Stuck at this point. Any more guidance you can give me?

    Here's a picture of the image:
    http://capa.physics.mcmaster.ca/figures/kn/Graph08/kn-pic0848.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  5. Jan 25, 2012 #4
    Well, where is the acceleration happening? over d1 or over d2?
     
  6. Jan 25, 2012 #5
    Okay, so acceleration is happening over d1.
    So, as he decreases his height by 61.2 cm. Using this distance in a=v2-v02/2d gave me an acceleration of 3.06 m/s^2, which I got wrong.
    Sorry, I'm not catching onto a major concept here...
     
  7. Jan 25, 2012 #6
    There's some algebra mistakes in there (unless /2d means [itex] \frac{v^{2} - v_{0}^{2}}{2d}[/itex]) . Firstly remember that v_o = 0. He isn't moving initially. V is the final velocity at take-off. Also don't forget to convert your cm to m! Edit -and also, v is squared!!

    edit2 -

    I'm not sure if this is a typo or a mis-understanding; but it's crucial to be able to understand what's happening step by step. The problem begins when he is in the crouched position. He then uses is legs to accelerate constantly to a certain height where his feet leave the ground, he is also at a certain velocity at this point. Then the rest is gravity.
     
    Last edited: Jan 25, 2012
  8. Jan 25, 2012 #7
    Notice it says BEFORE his feet leave the ground so look at picture number one and label your velocities for that picture, not for the part where he's flying through the air.
     
  9. Jan 25, 2012 #8
    Right - check your math and your directions so your signs are correct.

    In the first question he's flying through the air and slowing down. What slows him down? So this gives you your sign on your acceleration for the first question.

    But for the second question he's NOT slowing down so what does that mean regarding the sign of your acceleration?

    You have an algebra error or you typed it wrong when you solved for a.
     
  10. Jan 25, 2012 #9
    Thanks for your help! I had converted everything, it was the silliest mistake I was making, forgot to square the value of v.
     
  11. Jan 25, 2012 #10
    Glad you got it.
     
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