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Does this look right? Cannonball question

  1. Apr 7, 2005 #1
    Cannonball fired on 45 deg angle from horizontal. Initial position is origin. initial velocity 100sqrt(2) ft/sec. Trajectory is y=x-(x/25)^2 where y>=0.

    How far till it hits the ground? 312.5

    What's the max height? 156.25

    Here is what I got m(a) = 2*-(1/625)a + 1 = 0 which = -(1/312)a + 312 = 0

    -(1/625)*312.5^2+312.5 = 156.25

    Do you agree????
  2. jcsd
  3. Apr 7, 2005 #2
    Yep, I agree. :smile:
    Last edited: Apr 7, 2005
  4. Apr 7, 2005 #3
    Nope... his is right. He has the [itex]25[/itex] squared too~
  5. Apr 7, 2005 #4
    Argh, my brain just blurred over this one. :rofl:
  6. Apr 7, 2005 #5
    I made that mistake too, for a second, and was about to post it when I noticed he was right :tongue:
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