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Throwing a ball

  1. May 9, 2005 #1
    if I threw a ball straight up (including air friction), would it take longer for the ball to drop back to its original position from its max height or longer for the ball to rise up to its max height?

    here's what I'm thinking, on the way up, there is air friction and mg going agaist the initial velocity, on the way down, it would just be air friction against the ball. Since on the way down, there is only one resistive force, then it would fall faster right?

    also, air friction in this problem decreases the acceleration right?
     
    Last edited: May 9, 2005
  2. jcsd
  3. May 9, 2005 #2
    So mg doesnt act on the way down? Are you sure?
     
  4. May 9, 2005 #3
    the resistive force on the way up would be F=Fg+Ff

    on the way down it would be F=Ff

    since there is less resistive force on the way down, the ball would drop faster
     
  5. May 9, 2005 #4
    Net force on the way up, with up as positive:

    [tex] -(F_g + F_a) [/tex]

    Down:

    [tex] -F_g + F_a [/tex]

    Finding:

    [tex] -F_g - F_a (?) -F_g + F_a [/tex]

    Cancel -F_g

    [tex] -F_a (?) F_a [/tex]

    F_a is acting AGAINST the direction of motion in both directions. It's still my opinion that it takes the same time to go up and down.
     
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