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Distance traveled and hang time for 2D projectile

  1. Dec 4, 2012 #1
    1. The problem statement, all variables and given/known data

    A cannon is fired at an angle of 42 degrees, the cannon ball travels 8.9 meters in the x direction. Its hang time is 1.5 seconds. The cannon site .43 meters above the ground on a stand. If this same cannon was fired at 35 degrees, what would the hang time and the distance traveled be?

    2. Relevant equations



    3. The attempt at a solution
    I tried to use velocity vectors and find the speed at which the cannon ball was fired, muzzle velocity, using A^2+B^2=C^2. Giving me a velocity of 8.85m/s, using this I attempted to use sine and cosine to find the velocitys in the x and y direction. I am stuck here.
     
  2. jcsd
  3. Dec 4, 2012 #2

    haruspex

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    Must be something I'm not understanding. There seems to be too much information, and it conflicts. From the hang time (= time in the air, yes?) and the horizontal distance I can calculate horizontal speed. From that and the launch angle I can compute vertical speed at muzzle. From that, the hang time, and taking g=9.81 m/s2 I compute that the cannonball lands 3.02 m below the level of the muzzle, but we're told the cannon is only .43 m above the ground.
     
  4. Dec 4, 2012 #3

    CWatters

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    I agree...

    Vx = 8.9/1.5 = 5.933 m/s
    Then..
    Vx = V Cos(42)
    V = 7.984 m/s

    Vy = V Sin(42)
    = 5.342 m/s

    Vertically
    S=ut-0.5at2
    S=5.342 * 1.5 - 0.5*9.8*1.52

    = 8.013 - 11.025
    =-3m

    So it does appear to land 3m below the cannon. So perhaps the ground isn't flat?

    Perhaps an error in the height of the cannon?
     
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