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Man shot out of cannon question

  1. May 30, 2007 #1
    http://img263.imageshack.us/my.php?image=helpxy5.jpg
    (pic attached)

    Questions:
    How far did man travel?
    What was velocity upon hitting the wall? ( is it sufficent to cause death)

    Known:
    Time from cannon shot --> impact = 14 seconds ( on pic it says 16, but its actually 14)
    Angle of cannon = 36.86 degrees
    velocty = 30m/s

    2. Relevant equations

    vf^2 = vi^2+ viT ?

    3. The attempt at a solution

    i know that to find the distance traveled, is just the x-component.

    THerefore distance = 30cos(36.86)*14 = 335.42 feet???

    To find final velocity, I would think to use vf^2 = vi^2+ viT, but this is only for linear motion.

    Please help!
     
  2. jcsd
  3. May 30, 2007 #2

    when the man is going to land on the other building, he is going to land at angle thus he will have both X and Y components to his velocity. Since there is no force (thus no acceleration) in the X direction, the X component of his velocity will remain the same. Thus you must calculate the Y component of his velocity as well. Are the buildings the same height - thus what was his displacement in the Y direction? WHat is the acceleration i tneh Y direction only? What was his initial velocity? For how long was he in the air? Find the final velocity in the Y direction from this.

    now you can calculate his final velocity using both X and Y components
     
  4. May 31, 2007 #3
    https://www.physicsforums.com/newreply.php?do=newreply&noquote=1&p=1345358

    "Thus you must calculate the Y component of his velocity as well.
    Are the buildings the same height - thus what was his displacement in the Y direction? WHat is the acceleration i tneh Y direction only? What was his initial velocity? For how long was he in the air? Find the final velocity in the Y direction from this.

    now you can calculate his final velocity using both X and Y components"

    You do not know if the buildings are the same heights. There is also the chance, that he lands on top of the building.
    the Y displacement is 30sin(36.86)14 + 3 (the 3 is from the canon's height) - 4.9(14^2) = -693.88 feet.

    A = -9.81?
    Vi intial = 30 m/s or 0?
    air time = 14s
    using equation, vi^2 = 30^2 + 2(-9.81)*335 ( 335 is delta x) which equals
    -5672.7... which is not the right answer. can sombody verify the right answers T_T
     
  5. May 31, 2007 #4
    alright if you dont know the displacement, its fine

    you still know the initial velocity, the acceleration, the time of flight and thus you can find the final velocity

    you dont need to find the displacement to find the final velocity...
     
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