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Math Help - Probem Solcing (How Far can You See?)

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    Probem Solcing (How Far can You See?)

    Upper-story and penthouse apartments rent for the highest prices because of terrific views. Assuming the atmosphere is clear, and there are no hills or mountains obstructing the view, how far could you see to the horizon from the top of a 100 meter fire tower? From the top of 55 story skyscraper, where each floor is 3 meters high? From the top of the tallest building in the world (the Petronis Tower in Kuala Lumpur, Malaysia, 1,467 ft. high)? From an airplane flying at 35,000 feet? From the Space Shuttle flying at 300 km? (The radius of the earth if approximately 6378km or 3963 miles).
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    Quote Originally Posted by matgrl View Post
    Upper-story and penthouse apartments rent for the highest prices because of terrific views. Assuming the atmosphere is clear, and there are no hills or mountains obstructing the view, how far could you see to the horizon from the top of a 100 meter fire tower? From the top of 55 story skyscraper, where each floor is 3 meters high? From the top of the tallest building in the world (the Petronis Tower in Kuala Lumpur, Malaysia, 1,467 ft. high)? From an airplane flying at 35,000 feet? From the Space Shuttle flying at 300 km? (The radius of the earth if approximately 6378km or 3963 miles).
    Not sure of the easiest way, but you can find the desired tangent line using calculus.
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  3. #3
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    Hello, matgrl!

    Penthouse apartments rent for the highest prices because of terrific views.
    Assuming the atmosphere is clear and there are no obstructions,
    how far could you see to the horizon from the top of a 100-meter fire tower?

    From the top of 165-meter skyscraper?

    From the top of the tallest building in the world?
    . . (the Petronis Tower in Kuala Lumpur, Malaysia, 1,467 ft. high)

    From an airplane flying at 35,000 feet?

    From the Space Shuttle flying at 300 km?

    The radius of the earth if approximately 6378 km or 3963 miles.
    Code:
                    A
                    o
                    |\
                    | \
                   h|  \
                    |   \ d
                   C|    \
                  * o *   \
              *     |     *\
            *       |       o B
           *        |     *  *
                   r|   * r
          *         | *       *
          *         o         *
          *         O         *
    
           *                 *
            *               *
              *           *
                  * * *

    The center of the Earth is \,O.
    The radius is: \,r \:=\:OB = OC.

    The observer is at \,A.
    His altitude is: h \,=\,AC.

    He can see a distance of: d \:=\:AB.
    . . Note that \angle B = 90^o


    Pythagorus: . d^2 + r^2 \:=\:(h+r)^2 \quad\Rightarrow\quad d^2+r^2 \:=\:r^2 + 2rh + h^2

    . . . . . . . . . d^2 \:=\:2rh + h^2 \quad\Rightarrow\quad d \:=\:\sqrt{h(2r+h)}


    We can simplify further.

    Consider the factor (2r + h)

    Compared to the diameter of the Earth (2r),
    . . the height \,h is negligible.

    We can let: . 2r \:\approx\;2r+h


    The distance formula becomes: . \boxed{d \:\approx\:\sqrt{2rh}}


    Be sure to use the same units (feet/miles, meters/kilometers).
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    a "rule of thumb" ...

    line of sight range in nautical miles \approx 1.23\sqrt{h} , where  h is the altitude in feet.
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