How far away is the horizon when you're at the top of the Gateway Arch?
490274
What several answerers are failing to remember is that St. Louis is not at sea level.  The St. Louis Gateway Arch is 192 meters (630ft) high.  But it is sitting on land that is 132 meters (433ft) above sea level.  This comes to a total height above sea level of 324 meters (1066ft).\n\nI'll assume you are trying to figure out the straight line distance to the horizon (the point where a viewing line is tangent with the curve of the earth).  The curved distance along the surface of the earth is almost the same when close to the earth, so it really isn't worth noting, but if you were in a satellite, there would be notable difference.\n\nAnyway, using triangles, you are trying to find the leg of a right-triangle with sides r, r+h and d.\n\nTo derive a formula figure out that d^2 + r^2 = (r+h)^2.\n\nr = radius of the earth (6378.14km or 6,378,000m)\nh = height above sea level (324m)\n\nAnyway, plug this all in and you get that the distance to the horizon is 64.29km or about 40 miles.\n\n40 miles is the correct answer
10335
If we assume that visibility is unlimited, then the horizon is always the same distance away from a specific point. How far is the line of the horizon depends on the altitude of the point - the higher the point, the further away the horizon. By the same token, you will be able to see high mountains that are close to, but beyond the low-lying horizon line.\n\nOn other planets/moons the distance to the horizon is a function of their diameter - the bigger the body, the further away the horizon is from a given point.
1169146
At least 12 feet.\nI could be wrong though.
42704
20 cm unless you are near-sighted, in which case it is closer
703199
It is an exercise in elementary calculus to show that (ignoring air resistance) the elevation angle for maximum distance is 45 degrees.  The vertical velocity of the egg will be zero at the peak height, and will increase as v = gt^2/2 until it hits the ground.  Energy is conserved, so the vertical velocity when it strikes the ground will match the vertical velocity with which it was launched, and the horizontal velocity will be constant.   This is enough data to solve the whole thing, but there is a shortcut: in the parabolic path at issue, it is always true that the peak height is 1/4 of the horizontal distance.  Hence, the peak height here is 24.225 meters, and that makes it trivial to compute the time and hence the velocity.
566011
Varies, from about 10 ft in the bay areas, gradually increasing in height as you go further in land. The highest points reaching upwards 250ft
152957
It's really far away.
969415
It's a geometry problem.  Suppose you're high up in a building on the beach overlooking the ocean.  You're at height h above sea level.\n\nNow draw a circle which represents the earth, with radius r.  Your distance from the center of the earth is r+h.  The horizon is some distance away, x.  Your line of sight to the horizon is <tangent> to the radius of the earth, forming a right triangle, where the hypoteneuse is r+h.\n\nUsing the Pythagorean Theorem,\n\nr^2 + x^2 = (r+h)^2 = r^2 + 2rh + h^2\nx^2 = 2rh + h^2\nx = sqrt(2rh + h^2)\n\nThat's the exact answer,  But for practical purposes, 2rh is <much, much> bigger than h^2.  To see this, suppose you're 500 feet in the air.  That's about a tenth of a mile, and using miles, h^2 = 0.1^2 = 0.01.  On the other hand, the earth's radius is about 4000 miles, so 2rh = 2 x 4000 x 0.1 = 800.  This is 80,000 times bigger than h^2.\n\nSo we can safely ignore the h^2, and say x = sqrt(2rh).  That can be broken down further.  Again using miles (you can also do this in kilometers),\n\nx = sqrt(2rh) = sqrt(2r) sqrt h = sqrt(8000) sqrt h = 90 sqrt h\n\n(rounding off, since sqrt 8000 = 89.44; also, h must be in miles).\n\nUsing the formula x = 90 sqrt h, suppose you're in an airplane 4 miles up in the air (21,000 feet).  Then the horizon is 90 sqrt 4 = 180 miles away.\n\nOkay, now let's work it out for h in feet instead of miles.  We'll go back to the formula x = sqrt(2rh), and use r = 3960 miles (a more accurate figure).  Then\n\nx = sqrt(2 x 3960 x h / 5280) = sqrt(3h/2) = 1.225 sqrt h\n\nwhere h is in feet and x is in miles.\n\nTwo examples:  (1) For h = 6 feet, use x = sqrt(3h/2) = 3 miles.  (2) For x = 100 feet, use x = 1.225 sqrt h = 12.25 miles.\n\nTo recap:  For h in miles, use x = 90 sqrt h.  For h in feet, either use x = sqrt(3h/2) or x = 1.225 sqrt h.\n\nIn all cases, x will be in miles.  And of course, all of this can be done using meters and kilometers.
411381
About 6 feet, which isn't an issue, unless you live at,near or below sea level. If you do, you should move anyway.
385865
the height at which condensation takes place
384862
The air is thinner the higher you go.
311797
About 5 feet outward in all directions.
659467
i jumped right out of the stratosphere once and when i came down 1 hour later i was a thousand miles from my garden. by the time i got back home my wife had divorced me,remarried and had 10 more children. so please don't jump to high
507550
Wow... a lot of complicated answers...simply put... the majority of your force (or pressure) will be on the arch sideways with the most force on it at the initial take off point. It will gradually decrease as it goes round and up as gravity will take hold and it looses it momentum due to the friction against the arch(not allowing it to go straight. Remember that a body in motion wants to stay in motion. If the arch is not tight, it will carry most of its (the projectile) energy to the top platform, so make sure it is prepared for it.
1134341
The answer would depend upon the observer's height, or more accurately, the height of the observer's eyes from sea level.  For simplicity's sake, let's use the numbers 1.5m for eye height, 6,375,000m for the radius of Earth, and of course, assume that the planet is a perfect sphere, our observer has perfect eyesight, etc.\n\nThe line of the observer's vision when it meets the horizon forms a tangent to the surface of Earth.  If you draw a line from this point to the center of the planet, and another line from the observer to the center, you form a right triangle.  One leg is 6,375,000m and the hypotenuse is 6,375,001.5m.  So the other leg can easily be computed using the Pythagorean Theorem:\n\nd = sqrt[ (6,375,0001.5)^2 - (6,375,000)^2 ] = 4373.21\n\nSo to our observer, the horizon is about 4.37km away.\n\nNote that this is the limit of vision for objects with zero height, e.g. something painted on the ground (assuming again our observer has perfect eyesight).  S/he can see objects even further away if they do have height.  For example, two observers with identical height will be able to see one another at twice the distance, though it will only be one another's foreheads poking a tiny bit above the horizon.\n\nLarge objects such as buildings, or ships sailing in, may be seen at a much greater distance, and this distance is also made much greater if the observer moves to higher ground.  For example, even going to the top of a small, 10m building (only a single storey) will extend the horizon to 12.1km, and standing atop a 50m cliff extends it to 25.6km.
317223
It probably varies depending on where in the city you are, but it's around 1300 feet.
519171
Also launch at 45% in the direction of the wind, less height but more distance