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Acceleration (short question, but not so simple)

  1. Sep 7, 2005 #1

    This task have me puzzled:

    You throw a ball straight up in the air, and it reaches a lot higher than your own height. Is the size of the acceleration largest under the throw, or when the ball is moving downwards? Explain.

    It is strange, because I'm not sure that it gives enough information to make a plain answer... Anyone who can help?
  2. jcsd
  3. Sep 7, 2005 #2


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    Think of air resistance.
    Opposite the direction of motion.
  4. Sep 7, 2005 #3
    not sure

    I don't think that is what they are asking for, anyway it is the same air resistance both ways...
  5. Sep 7, 2005 #4


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    but when the ball is going up the air resistance is downward and the force on the ball is
    - mg - R
    and when the ball is coming down the air resistance is upward hence resultant force on the ball is - mg + R
  6. Sep 7, 2005 #5
    its easy...when the ball is thrown upwards the resultant force acting on it is the same as when is is falling down... meaning having the same acceleration...but due to gravity pull from the earth acceleration increases when it falls therefore the answer is.... the acceleration is faster when the ball falls
  7. Sep 7, 2005 #6
    therefore wat is ur answer???
  8. Sep 7, 2005 #7
    sorry mukundpa


    you are of course right, but I'm wondering if the question is about before the ball leaves the hand or just when it is on it's way upwards.

    Anyway: Thanks
  9. Sep 7, 2005 #8
    When the ball is released from your hand, the acceleration is constant until it hits the ground. As far as if it takes greater acceleration to throw the ball "a lot higher than your own height," that just depends on how tall you are and where you release it.
  10. Sep 7, 2005 #9


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    sorry, I thought it that way. Can't get meaning of under the through.
  11. Sep 8, 2005 #10


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    The key words are that the ball reaches a lot higher than your own height.
    The ball reaches an upward velocity during you keep it. Its displacement can not be longer than twice the length of your arm, and that is a bit shorter than your height. It accelerates form zero velocity to v during a displacement h. You know the relation between acceleration, displacement and final velocity?
    When you release the ball it has got the upward initial velocity v and it rises to height H, when its velocity becomes 0. You can use the same formula among magnitude of acceleration (it is g now) height and initial velocity that you used before for the final velocity.
    H >>>h. Can you go ahead?
  12. Sep 8, 2005 #11
    Have I understood it correctly?

    You are refering to: v=v0+2as

    So if I say my arm is 0.70m long, and the ball reaches 5 meters up in the air, I find that the acceleration during the throw is ca. 3.5 m/s^2.

    And therefore, if one doesn't include air resistance, the acceleration is larger when the ball falls freely (a=g=9.81).

    Could this be correct?
  13. Sep 8, 2005 #12


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    No. Your formula is wrong. Just check the units.

  14. Sep 9, 2005 #13
    Of course


    I see that I have forgotten to put the speed in second (v^2), and I actually think I can evaluate the whole question now!

    Since the distance the ball travels to gain the speed v is shorter than the distance the ball travels in the air to gain the same speed, the acceleration must be higher during the throw!!!
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