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Assistance with projectile motion question

  1. Apr 9, 2012 #1
    I had recently received an exam back rom my kinematics unit in my physics course, this is the only question that I answered and was incorrect. I was wondering if someone could assist me on what steps I am missing. The correct answer is 3.2m, but I can't figure out how to get to it.


    A ball is launched from a cliff with and initial velocity of 15 m/s at an angle of 40o above horizontal. If the ball lands 26 m away, determine the height of the cliff.

    Horizontal

    V=cos40(15)
    v=11.5

    T=d/v
    T=26/11.5
    T=2.26s

    Vertical

    V1=sin40(15)
    v1=9.64

    d=v1t+1/2at^2
    d=9.64(2.26)+1/2(-9.8)t^2
    d=-2.5m

    Any feedback would be greatly appreciated.
     
  2. jcsd
  3. Apr 9, 2012 #2
    You made some kind of simple arithmetic mistake in the final equation.

    9.64(2.26)+1/2(-9.8)(2.26)2≠-2.5
     
  4. Apr 9, 2012 #3
    I get -3.24 for this calculation using t=2.26
     
  5. Apr 9, 2012 #4
    I figured out that it was a matter of simply not taking my time. Thanks for the help, greatly appreciated.
     
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