Hello, strummer!

A sailor at the seashore watches a ship with a smokestack 30 meters above water level

as the ship steams out to sea. . The sailor's eye level is 4 meters above water level.

About how far is the ship from shore when the stack disappears from the sailor's view?

(The radius of Earth is about 6400 kilometers.) Code:

A C B
o - - - - * o * - - - - o
4\ * | * /30
* | *
* \ |R / *
R \ | / R
* \ | / *
* o *
* O *
* *
* *
* *
* * *

The sailor is at $\displaystyle A$, the top of the smokestack is $\displaystyle B.$

The line of sight is tangent to the Earth at $\displaystyle C.$

The center of Earth is $\displaystyle O$, its radius is $\displaystyle R$ (6,400,000 m).

In right triangle $\displaystyle OCA\!:\;\;AC^2 + R^2 \:=\:(R+4)^2 \quad\Rightarrow\quad AC \:=\:\sqrt{8R + 16}$

In right triangle $\displaystyle OCB\!:\;\;CB^2 + R^2\:=\: (R+30)^2\quad\Rightarrow\quad CB \:=\:\sqrt{60R + 900}$

Hence: .$\displaystyle AB \:=\:AC + CB \:=\:\sqrt{8R + 16} + \sqrt{60R + 900}$

. . . . . . . . . $\displaystyle = \;\sqrt{8(6,\!400,\!000) + 16} + \sqrt{60(6,\!400,\!000) + 900} $

. . . . . . . . . $\displaystyle = \;26,\!751.35955$ m.

Therefore, the distance is about 26.75 kilometers.

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Edit . . .

Ah, I see that skeeter did it *correctly* (ship-to-shore distance).

Using the Law of Cosines, $\displaystyle \theta = \angle AOB$ is given by:

. . $\displaystyle \cos\theta \;=\;\frac{6400.004^2 + 6400.03^2 - 26.75^2}{2(6400.004)(6400.03) } \:=\:0.999991264$

Hence: .$\displaystyle \theta \:=\:0.00417989$ radians.

Therefore: .$\displaystyle \overline{ACB} \;=\;r\theta \;=\;(6400)(0.00417989) \;=\;26.751296 \;\approx\;26.75$ km