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At what angle should the bomb be released?

  1. Nov 23, 2015 #1
    1. The problem statement, all variables and given/known data
    There is a plane flying at 200 km/h and wants to drop a bomb on a car traveling 130 km/h which is 78.0m below the plane. at what angle should the bomb be released? the answer is 45 degrees.

    2. Relevant equations
    sine law: SinA/a= SinB/b
    cosine law: a^2=b^2+c^2+bcCosA
    v=d/t
    sine,cosine and tan trig ratios

    3. The attempt at a solution
    converted 78.0m to 0.078 km.
    I have drawn out the vectors but have not been able to add them. Any help would be appreciated.
     
  2. jcsd
  3. Nov 23, 2015 #2

    SammyS

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    It seems to me that there is not sufficient information to answer this.

    Have you stated the entire problem?

    Is there a figure to go with this?
     
  4. Nov 23, 2015 #3
     
  5. Nov 23, 2015 #4

    SammyS

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    Do you know the relative positions of the plane and the car, as well as the relative directions of motion?
     
  6. Nov 24, 2015 #5

    CWatters

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    I agree with your answer.

    However I do think the problem statement is badly worded - for example it says the car is "below" the aircraft which isn't the case. The answer also assumes the aircraft and car are going in the same direction. If they are going in opposite directions you get a different answer.
     
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