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Ship's displacement vector

  1. Aug 24, 2006 #1
    Hi, I am new here and I am taking Ap Physic I have some problems and I can't figure them out I posted them below. Any help would be appriciated...:rofl:

    You are driving into St. Louis, Missouri, and in the distance you see the famous Gateway-to-the-West arch. This monument rises to a height of 192 m. You estimate your line of sight with the top of the arch to be 7.8° above the horizontal. Approximately how far (in kilometers) are you from the base of the arch?_____km

    An ocean liner leaves New York City and travels 47.0° north of east for 218 km. How far east and how far north has it gone? In other words, what are the magnitudes of the components of the ship's displacement vector in the following directions?
    (a) due east ______km
    (b) due north______km
     
  2. jcsd
  3. Aug 24, 2006 #2
    Hi, Tarylor. Welcome the forum.
    You have to show us what you have done before asking for help. Both the answers require nothing more than simple trigonometry.
     
  4. Aug 24, 2006 #3
    I dont really know how to show you what I've done cause it is a drawing I just need instructions on how to solve
     
  5. Aug 24, 2006 #4
    For the first, use the definition of the tangent of an angle. The second requires to use the sine and cosine functions.
     
  6. Aug 24, 2006 #5
    K

    K I got this one can you help me with this one.

    You are driving into St. Louis, Missouri, and in the distance you see the famous Gateway-to-the-West arch. This monument rises to a height of 192 m. You estimate your line of sight with the top of the arch to be 7.8° above the horizontal. Approximately how far (in kilometers) are you from the base of the arch? _______km
     
  7. Aug 24, 2006 #6
    I assume that you have drawn the appropriate diagram for this problem. Do you know that the ratio between the height of the monument and your distance to it is?
     
  8. Aug 24, 2006 #7
    actually i was unsure if this looked like a right triangle or if it had an arc
     
  9. Aug 24, 2006 #8
    For problems such as these, you can safely assume that it's a right triangle, unless otherwise mentioned, of course.
     
  10. Aug 24, 2006 #9
    ok i made the triangle and it looks like this

    [​IMG]
     
  11. Aug 25, 2006 #10

    andrevdh

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    You can now use the tangent of the angle above the horizontal to calculate the required distance. I've checked in the internet and the arc's height is 630 feet.
     
  12. Aug 25, 2006 #11
    Btw, the height of the arch is 192 metres, not Km.
     
    Last edited: Aug 25, 2006
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