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Motion in 2d question

  1. Sep 20, 2007 #1
    1. The problem statement, all variables and given/known data

    A dive-bomber has a velocity of 280 m/s at an angle of [tex]\vartheta[/tex] below the horizontal. When the altitude of the aircraft is 2.15km it releases a bomb, which subsequently hits a target on the ground. The magnitude of the displacement from the point of release of the bomb to the target is 3.25km. Find the angle [tex]\vartheta[/tex].

    3. The attempt at a solution

    I am really lost with this question, so if some one could give me some clues that would be wonderful. I also don't really know what the question looks like, so its hard to know where to start.

    Thanks for the help!
     
  2. jcsd
  3. Sep 20, 2007 #2

    cristo

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    Staff Emeritus
    Science Advisor

    Try drawing a diagram to set the problem up.
     
  4. Sep 20, 2007 #3
    I get a major brain-fart everytime I try to draw it out. I dont know to even draw it out.

    The teacher also gave the hints.

    1) Find the horizontal distance x
    2) Obtain the expression for y as a function of x for all angles [tex]\vartheta[/tex].
    3) Find [tex]\vartheta[/tex] by equating y(x=xf = -h, and using the trigonometric relation.

    1 = 1 + tan^2 [tex]\vartheta[/tex]
    cos^2 [tex]\vartheta[/tex]
     
    Last edited: Sep 20, 2007
  5. Sep 21, 2007 #4

    learningphysics

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    Homework Helper

    Write the equation for vertical displacement in terms of time.

    Write the equation for horizontal displacement in terms of time.
     
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