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Another kinematics problem

  1. Sep 11, 2010 #1
    1. The problem statement, all variables and given/known data
    A pilot flies horizontally at 1300 km/hr at height 35m above initially level ground. At t= 0 the pilot begins to fly over ground sloping upward at an angle of 4.3 degrees. If the pilot does not change his heading, at what time will he hit the ground?


    2. Relevant equations

    Kinematics equations, and maybe trig.

    3. The attempt at a solution

    I have two ideas on how to start this problem. First was to think of the plane as going down at the angle of 4.3 degrees and the ground being flat and draw a triangle. But I felt like I was on the wrong track.
    Then I was thinking to think of the airplane as sitting still, and the ground coming up toward it a 4.3 degrees and 1300 km/hr.

    Are either of these the right way to solve this problem?
    Thanks
     
  2. jcsd
  3. Sep 11, 2010 #2

    kuruman

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    You can try solving the way you suggest, but you may get confused. There is more straightforward way. Draw yourself a right triangle with one side vertical and equal to 35 m and with the angle opposite to it equal to 4.3 degrees. Can you calculate the length of the other right side that is horizontal? If so, find the time it takes the plane to travel that horizontal distance.
     
  4. Sep 11, 2010 #3

    lewando

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    Always a good idea to draw a picture exactly related to the question. You can always take shortcuts later after youv'e done infinity of these types of problems. I'm imagining a right triangle in your future with one of the three angles equal to 90-4.3 degrees. You should be able to find the distance the plane will travel using trig, as you correctly asserted. Then its a matter of using the constant velocity equation d=r*t.

    [edit: I don't even have a cool picture. I'll go to bed now.]
     
  5. Sep 12, 2010 #4
    2m6sqar.jpg

    My trig is a little rusty, but I think I can use the law of sines to get the horizontal magnitude.

    [tex]\frac{L}{sin85.7}=\frac{35}{sin4.3}[/tex]

    [tex]L\approx470m[/tex]

    The plane is moving at 361m/s, so

    [tex]\frac{470m}{361m/s}\approx1.3s[/tex]

    Does that look right?
     
  6. Sep 12, 2010 #5

    lewando

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    Yes. Good job.
     
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