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Circular Launch

  1. Sep 7, 2009 #1
    A ball is launched up a semicircular chute in such a way that at the top of the chute, just before it goes into free fall, the ball has a centripetal acceleration of magnitude 2 g.

    How far from the bottom of the chute does the ball land?

    Height=2R

    I got initial velocity=sqrt(2gR), and I plugged that into d=vit+1/2at^2, to give me 2R=sqrt(2gR)t+1/2gt^2, and I used the quadratic formula and got [-sqrt(2gR)+sqrt((2gR)-4gR))]/g, which simplifies to [sqrt(6gR)-sqrt(2gR)]/g. But, that was wrong...and I'm really not sure where I went wrong. Can someone tell me what the problem is here?
     
  2. jcsd
  3. Sep 7, 2009 #2
    distance of downward fall = 2R. time of fall = sqrt(4R/g) from d = 1/2 g t^2. Note: the
    initial vertical velocity is zero. (I am thinking of the chute as the right half of a circle---I hope this is the right picture).

    Hence horizontal distance traveled = the initial velocity you got, namely sqrt(2 g R),
    times the time of fall. This gives horizontal distance traveled = sqrt(8)*R or 2 sqrt(2) R.
     
  4. Sep 7, 2009 #3
    Thank you, I thought the vertical velocity was sqrt(2gR) for some reason. I see it now, thanks.
     
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