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Physics velocity problem

  1. Feb 17, 2013 #1
    1. The problem statement, all variables and given/known data A man is going to sail his sailboat across a bay to a harbor from his dock and wants to know how long it will take him. He knows that if he sails directly across the bay from his dock he will reach the harbor. However, he has no charts so he doesn't know the distance to the harbor. There is a tower at right angles the line between the dock and the harbor. The tower is 3.75x10^3 meters from the dock and is 10^2 meters high. He gets into his car and drives to the tower. He climbs 50 meters up the tower until he can see the harbor. He measures the angle between his line of sight to the harbor and the direction of the dock to be 79.64 degrees. He estimates that his boat can maintain a velocity in the direction of the bow of 100 meters per minute with the wind that is blowing. How long will it take him in hours and minutes to sail across the bay to the harbor.



    2. Relevant equations

    Law of Cosines, Law of sines, T=D/V



    3. The attempt at a solution
    I've attempted to draw the picture, but I know my drawings are incorrect because my answers come out to be negative somehow. If someone could just draw the picture for me, I could go on from there
     
    Last edited: Feb 17, 2013
  2. jcsd
  3. Feb 17, 2013 #2

    haruspex

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    Pls post your drawing first.
     
  4. Feb 17, 2013 #3
    ImageUploadedByPhysics Forums1361164576.005995.jpg - That is my most recent drawing. I believe it is correct, but I don't know where to include the 79.64 degree angle
     
  5. Feb 17, 2013 #4

    haruspex

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    You need to represent the third dimension, with a perspective drawing or several plane drawings. E.g. you could draw one for the plane containing the dock, D, the harbour, H, and the viewing point from the tower, V.
    Can you work out the distance DV and the angle VDH?
     
  6. Feb 17, 2013 #5
    I got DV to equal 4790.228253 and angle vdh to be simply 45 degrees
     
  7. Feb 18, 2013 #6

    haruspex

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    The question is not worded very well. It does not make it clear where the tower is in relation to the dock. Saying it's "at right angles to the line ..." doesn't mean anything unless the point of reference on the line is given. The only interpretation that makes sense to me is that BDH is a right angle, where the base of the tower is point B.
    On that understanding, what is angle VDH?
     
  8. Feb 18, 2013 #7
    ImageUploadedByPhysics Forums1361246196.103417.jpg I managed I figure out. You're right it's not worded well enough and that's what really confused me. Apparently all I had to was put 79.64 as the angle on the tower.
     
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