An offshore drilling rig is being towed out to sea. What is the maximum distance away that the navigation lights can still be seen by an observer standing at the shoreline? The observer's eye height is 5'9" and the uppermost navigation light is 225 ft above the water.
I need some help solving this problem. Please show steps. Thanks!!!
icprofit6000
Assuming a circular earth with radius r (in feet).
form a triangle with a point on the earth, a line tangent (at right angles) to a point which is r+225ft in altitude, the distance to this point would be sqrt( (r+225)^2-r^2)
from the point on the earth form another right triangle to the observer distance would be sqrt( (r+5+9/12)^2-r^2))
Add the two distances.
Of course, this assumes that light travels in a straight line and ignores scattering and transmission lost (e.g. there no fog)
The tangent point is the farest point on the earth that the observer can see. To see a point farther it much be above the earth.
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