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Easy questions, A ball is shot out from a launcher, what is the acceleration?

  1. Sep 23, 2012 #1
    A ball is shot out from a launcher as shown in the figure.

    http://insight.ccsf.edu/file.php/4576/Lab_7_Prelab_Q1.JPG
    (The picture is a ball being shot from a canon from on top of a building going down)

    The acceleration of the ball is

    a. zero
    b. constant
    c. increasing continuously
    d. decreasing continuously

    i just want to make sure, the answer is a) right??
     
  2. jcsd
  3. Sep 23, 2012 #2

    PhanthomJay

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    I can't open the pic, but once an object leaves the cannon, and in the absence of air resistance, it is acted on by the force of gravity only. Why do you say the acceleration is 0?
     
  4. Sep 24, 2012 #3
    accel. is zero because since only the force of gravity is acting on it then it does not speed up or slow down so it is zero. am i right??
     
  5. Sep 24, 2012 #4
    I cant see the picture as you need a log on but,
    acceleration due to gravity is 9.81ms-2

    therefore it must be increasing constantly as it falls, when the ball reaches maximum height acceleration is 0 for an instant
     
  6. Sep 24, 2012 #5

    PhanthomJay

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    No. If an object is subject to a net force, it must accelerate in the direction of the net force, per Newton's 2nd law, F_net = ma. The net force here is the object's weight , acting down, mg, where m is the mass of the object and g is the acceleration of gravity. So mg = ma. Is the acceleration, a, constant, increasing, or decreasing?
     
  7. Sep 25, 2012 #6
    i chose the answer is c. increasing continuously, but it is incorrect...so im leaning more towards accel is either constant or zero. The ball is shot horizontally by the way.
     
  8. Sep 25, 2012 #7
    Why do you think its c? The force acting on the ball is always the force of gravity (even if its shot horizontally) which is constant.
     
  9. Sep 25, 2012 #8
    im just listening to what Johnahh said "therefore it must be increasing constantly as it falls, when the ball reaches maximum height acceleration is 0 for an instant"
     
  10. Sep 25, 2012 #9
    Ask yourself, is that correct? Read carefully PhantomJay's post.
     
  11. Sep 25, 2012 #10
    ah i see since mg = ma then it should be constant right?
     
  12. Sep 25, 2012 #11
    What do you think? If i answer that, i solve your homework.
     
  13. Sep 25, 2012 #12
    thanks pranav!
     
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