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Throwing and dropping something

  1. Mar 1, 2012 #1
    When I throw a ball up in the air, my acceleration will still be -10m/s^2 right? Just that my original upwards force before I release it will result in a greater upwards speed, then when the hand stops producing the force, the weight will result in a -10m/s^2 causing a deceleration and eventual a downwards acceleration of 10m/s^2 right?

    Then when I throw a ball down from a building. In addition to the force acting downwards there is weight, so at the point my acceleration is more than 10m/s^2. Then when I release the ball, I have a decreased force. So will the acceleration start to decrease to 10m/s^2? So that means I will increase in speed but now in a slower rate as the ball goes down?

    Thanks for the help!
     
  2. jcsd
  3. Mar 1, 2012 #2

    Doc Al

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    Staff: Mentor

    While the ball is in your hand, there are two forces acting on the ball: the force from your hand and gravity. As soon as the ball leaves your hand, the only force acting is gravity (ignoring air resistance) which gives a constant acceleration of 9.8 m/s^2 downward. That acceleration is the same regardless of whether you drop the ball, throw it up, or throw it down.
     
  4. Mar 1, 2012 #3
    All the time that your hand is holding the ball, it could be accelerating at whatever rate your muscles are capable of making it. The rate of acceleration could be changing. At the moment it leaves your hand, the only thing that is important is its velocity: if we know its speed and direction of motion at this moment, we can calculate its trajectory.

    As soon as it leaves your hand, the only thing acting on it is gravity, which gives it a constant downwards acceleration of 9.8 m/s^2.
     
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