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Determining angle from initial velocity and distance

  1. Nov 4, 2008 #1
    1. The problem statement, all variables and given/known data

    A gun fires a round that goes 122 km in 170 seconds before impacting the ground. Ignore air resistance and the curvature of the earth. Find the muzzle velocity and angle.

    2. Relevant equations

    Vx = Vi(cos(theta)) Yx = Vi(sin(theta)) t= (-2Vy/-9.8) and D=(Vi2(sin(2theta)))/-9.8

    3. The attempt at a solution

    I found the correct muzzle velocity. Use distance 122*1000 then divide by time. This is horizontal velocity 717.647. Then use the third equation to find vertical. 170 = -2Vy/9.8.

    Then use pythagorean to find final vector: 1099.502752

    Now for the angle I used the fourth equation.

    122000 = 1099.5027522(sin(2theta))/9.8

    .9889931081 = sin(2theta)

    40.74559192 degrees I guess this answer is wrong. I know the muzzle velocity is correct but according to the teacher it the angle is incorrect. Are my equations flawed or did I mess up?
     
  2. jcsd
  3. Nov 5, 2008 #2

    Redbelly98

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    The problem with the method you used is, there are two different angles between 0 and 90 where sine(2theta)=0.98899...

    How using Vx and Vy to draw a right triangle, then use trig to get the angle?
     
  4. Nov 6, 2008 #3
    I got it. I used the Vx velocity which was easy to find then found the height assuming that it is landing at the same height and used tangent to find the final velocity vector. Thank you.
     
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