DISTANCE between the beach and HORIZON// how do you do it?
931645
first answer was close to right.\n\nthere is one qualification that seems need.\n(do this all in meters for easyness of calculation)\n\nR = radius of earth = 6400 km = 6.4 * 10^6 meters\nd = distance to horizon in meters\nh = height in meters\n\nSimilar triangles:\n\nR / d  = d / 2h\n\nd^2 = 2Rh\n\nd = sqrt (2Rh)\n\n....  works pretty well for small "h"\n\neg 2 meters height =>\n\nd = sqrt (2 * 6.4 * 10^6 * 2 ) = sqrt (25.6 * 10^6) \n=5.06 km\n\n[[ online calculator gives 5.5 km ]\n\nmine's a LOT easier to use ...\n\nI use "2h" instead of "h" for similar triangles\nbecause of discrepancy of right angle at the\nradius from eearth-center to observer and\nr-a from earth-center to distant object.\n[intersection happens "h" below earth surface]\n\nthat sounds confusing but I can't draw a picture\n\nmy eqn is easiest possible tho ..
725888
Using Pythagoras' Theorem, with R, the radius of the earth, equal to 6400km, and assuming that your height is considerably less than R, then you find that the distance to the horizon is approximately 3.6km multiplied by the square root of the height above ground level.\n\nFor example, if you are roughly 2m tall then the distance, if you are standing at ground level will be 3.6x1.4 = 5km, roughly.  This is about 3 miles.\n\nIf you are 2m tall and standing on a cliff of 20m then you can see 3.6x4.7 = 17km or about 10.5 miles.\n\nHope this helps!
1084738
I'm not certain, since it wouldn't be a straight line.  A straight line from Florida would put you off the coast of Australia:\nhttp://map.pequenopolis.com/\n\nHope this helps!
1047992
think about looking at the horizon, its a line that runs from side to side , sky line in the horizon(horizontal line)
179969
Up to the horizon...!
20069
It depends a lot on how high above the water your eyes are and how tall the distant object is.\n\nThe radius of the Earth at the equator is 6378135 meters.  Assume your eyes are 2 meters above sea-level, and the distant object is at sea-level.\n\nYou have a right triangle, where the longer leg is the radius of the Earth, and the hypotenuse is 6378315 + 2.  The shorter leg is the distance to the horizon.\n\nThe distance to the horizon = √(6378317^2 - 6378315^2) = 5051 meters, or about 3 miles.\n\n\nIf the distant object is 2 meters above sea-level, then it'll disappear when it's another 3 miles past the horizon, or 6 miles from you.
523034
The first answer *edit* was modified so now what I say here makes no sense - but her website confirms that I am right (5km = ~3 miles)*. If we want to know the distance to the horizon from a vantage point on a sphere, the only thing that matters is the radius of the sphere and the height of the vantage point.\n\nSo, assume you are standing with your eyes at 5'5" looking out at a calm ocean. You will typically be able to view the sea surface at 3 arcminutes, or roughly 5 kilometers, out to sea. Beyond that you would still be able to see ships and whatever else sticking above the water quite easily, but upon close inspection realise they appear deeper in the water than they actually are.\n\nAlso note that light bends around mass, so you might be able to "see" farther than your actual vantage point, but I'm not doing that calculation.\n\nI assume that people who have values higher than mine are thinking about higher vantage points than mine, because I think my estimations are close and I'm positive that my logic is right. Correct me if my numbers are off, but I'd like to know why.
130197
Off the shore is offshore.  If you are leaving the US and trying to get into international waters, the distance is 2 miles.
733568
there are only three thing you need to get a great view.  1 make sure your in an area with no and I meen no man made light, 2 any field that is at least 1 acer this is do to the view of the event horizon, 3 last but not least, the higher up you are from sea level the better,  oh and if you can try to stay behind the wind if clouds are going to move in you want to be in a spot that the wind is not going to push then over head.  with that in mind you should have one heck of a show.  Have fun!!!
189082
If this isn't a math question and you don't want to buy a tool and you don't have a protractor, there should be a leveled board on the ground, and your tilted board should be long enough to reach it. Mark the ground board where the tilted board touches it. Mark a point on the tilted board (the higher it is the more accurate your answer will be), drop a plumb line from there to the ground board, and mark where it touches. Measure the length from the bottom of the tilted board to the upper plumb-line mark (call it "A") and the distance between the two marks on the ground board (call that "B"). The cosine of the angle you want is B/A. Many calculators will give you the angle (using an arc-cosine function).
767544
Let r=the earth's radius and h=your height.\n\nDraw a right triangle ABC, where AC = length r, BC = length r+h, and <A is the right angle and angle C is very small.  C is the center of the earth, and BA is tangent to the earth.  BA is the distance from your head to the horizon.  By Pythagoreas' theorem, \n\nlength(BA) = sqrt((r+h)^2 - r^2)\n= sqrt(2rh + h^2).
821134
One does not dive in a straight line, unless one is diving straight down. The effect of gravity causes the body to accelerate in the vertical direction, while (ignoring wind resistance and the curvature of the earth) one travels at a constant speed in the horizontal direction.\n\nSo this become a question of the length of the parabolic curve that you follow in travelling from the diving block to the water. This will be slightly greater than the 5-foot straight-line distance to the landing point (the splash point).\n\nIf you are studying calculus, write an expression for the path that the body follows, and apply the following formula: \ndistance = Integral (sqrt(1 + (dy/dx)^2) dx)\n\nRegarding your followup question:\nIf you're assuming that everything is a straight line, you just ratio everything up (using the concept of similar triangles), so the answer is 6/5 of 3 feet.\nIf you're using calculus, you need to use the concepts in the explanation above, and eventually use algebra to solve for the height of the platform.
540028
Pacific, by a wide margain.\n\nPacific Ocean Coastline: 7,623 miles\nAtlantic Ocean Coastline: 2,069 miles
82755
six miles OFF SHORE??
735157
let x be the distance AP and let Y be the angle <PLA.\n\nThen x=3tanY then differentiate implicitly to time\n\ndx/dt= 3sec^2Y dY/dt\n\nNow dY/dt= 4 rev/min * 2pi radians/rev= 8pi radians/min\n\nWhen x=1, we have tanY= 1/3 sec^2Y=1+tan^2Y= 10/9\n\nThus dx/dt=3*10/9*8pi= 80/3 pi which approximately equals\n\n83.8\n\nThe spot of light is moving along the shoreline at a rate of about \n\n83.8km/min when it is1km from p.\n\n*It is essential to convert revolutions to radians
571553
good question. depends on how far away it is. if the horizon is flat and you can see 10 miles away then it would be like a circle with a 20 mile radius...becuase it goes all the way around you...get it?
786773
Their distance from the Sun