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Force and Simple Dynamics: Dropped Tennis Ball

  1. Apr 11, 2005 #1
    A 0.3 kg tennis ball is dropped from rest at a height of 3.6 m onto a hard floor.
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    a) What is the speed of the ball at the instant of contact with the floor?
    |v| = m/s *
    sqrt( 2*9.81*3.6) OK

    HELP: This is just a problem in one-dimensional kinematics with constant acceleration. What is the acceleration of the ball while it is in the air?


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    A flash photograph shows that the ball is compressed a maximum of 0.6 cm when it strikes the floor.
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    b) Assuming that the acceleration of the ball is constant during its contact with the floor, what force does the floor exert on the ball?
    |f| = N

    HELP: Again, this is a problem in one-dimensional kinematics with constant acceleration. What is the acceleration of the ball when it is in contact with the floor?


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    c) Over what time does the force act in bringing the ball to rest?
    D t = s

    Request some guidance as how to proceed with part b and c.

    Thanks,
     
  2. jcsd
  3. Apr 11, 2005 #2

    dextercioby

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    Science Advisor
    Homework Helper

    What is the only force acting on the ball,while in the air.What is the force acting on the ball,while in contact with the ground?

    What it the definition of average force?

    Daniel.
     
  4. Apr 16, 2005 #3
    Force due to gravity while in air , and on contact the balls acceleration.

    Ft = mv

    know m , v ... but dont know time
     
  5. Apr 16, 2005 #4
    I know part c. once I find part b.
     
  6. Apr 16, 2005 #5
    You know the kinetic energy of the ball when it is about to hit the ground.

    You know that this energy was countered by the ground with a constant force that lasted x = 0.6cm.

    You can find the force :)
     
  7. Apr 16, 2005 #6
    Well, the force responsible for the moton is gravity, so the acceleration is g

    [tex]W = E_k^{END} - E_k^{BEGINNING}[/tex]

    W is work and in this case equal to 0.6*F and the change in kinetic energy is just the transferred kinetic energy when the ball hits the gorund (this is easy to calculate)...Solve this equation for F. the question really is : do you understand what this formula means physically ? Do you know what is happening ?

    You can find this yourself, like you said

    marlon
     
  8. Apr 16, 2005 #7
    Careful, 0.6*F isnt the correct equation. The ball was compressed 0.6cm, so 0.006m*F.
     
  9. Apr 16, 2005 #8
    Thanks for the correction whozum

    marlon
     
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