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Jumping from a roof

  1. Dec 11, 2009 #1
    1. The problem statement, all variables and given/known data
    A person jumps from the roof of a house 4.5m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70m. This the mass of his torso(excluding legs) is 45Kg,find (a) his velocity just before his feet strike the ground and (b) the average force exerted on his torso by his legs during deceleration.

    2. Relevant Questions
    I really dont know how to start this question and all i really got was Force of gravity
    and i dont even know if its used in this question.
  2. jcsd
  3. Dec 11, 2009 #2
    this is very similar to the other problem you posted, only in reverse.

    Since it asks for velocity before he hits the ground--how can we calculate this?

    We just equate the potential energy on the roof with the kinetic energy as he strikes the ground:

    mgy=1/2mV^2. We can proceed using the approach of the last problem where

    F the force of deceleration x distance(d) is equal to mgy (or 1/2mv^2 since they are the same), or using kinematics and looking only at velocity:

    V^2=2(a)*d where d is the 0.7m.
  4. Dec 11, 2009 #3
    what does mgy mean?
  5. Dec 11, 2009 #4
    Sorry mgy=mass*gravity*displacement above ground.
  6. Dec 11, 2009 #5
    i got his velocity,but now i need the average force exerted on his torso by his legs during decelertation
    Would A =9.8?
    I thought it would go like this
    Where Fg=441 and mass= 45 and acceleration=9.8
    but that just gives me a Air resistance of 1 and i dont think thats right
  7. Dec 11, 2009 #6
    it is not air resistance which is slowing his descent--it is from the stiffness of the springs (muscles) in his legs--go out and jump off a 3 foot wall. How do you land? With your legs slightly bent and then allow the angle in the knees to decrease so he is in a near squat when his center of mass comes to rest--this is not the same as his feet coming to rest. If this isn't clear, go out and try it!

    So what we need is a deceleration that will bring his body (center of mass) to rest in 0.7 meters--distance his chest drops as he "lands".

    2a*y=V^2 where v is the velocity at which he hits the ground. Solve for a.

    Since f=ma, calculate force.

    EDIT: y= 0.7 m if this isn't clear, not the distance off the roof.
  8. Dec 11, 2009 #7
    I now got it!!
    you do not know how much i appreciated you helping me through this!!
    Thank you!!!
  9. Dec 11, 2009 #8
    Cool, and you are most welcome.
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