A ball is tied to a string a swung around...how long must string be if you want ball to be speed of light?
814844
The length of the string is irrevelent unless the speed of rotation (rpm) is a constant. Theoretically, you could increase the rpm's on an infinately short string and the ball would reach the speed of light. However, if the rpm's were held at a constant, the speed of the ball through space would increase as the length of the string increased. However, if the constant rpm were at a minimum, the length of the string would have to be near infinity in order for the ball to reach the speed of light.
1150360
The acceleration due to gravity is 32.2 ft. per second per second, which is pretty near 32.  Now we need two fundamental equations of physics:\n\nv = a / t\n\nwhich we can re-write\n\nt = a / v\n\nSince a (acceleration) is 32 ft/sec^2, and v (velocity) is 32 ft/sec, time (t) is 32/32 = 1 second until the ball's velocity equals zero.\n\nSo how much distance in 1 second? The second equation gives us the answer:\n\nd = 1/2 a t^2\n\nwe know a = 32, and since we now know t = 1, we can solve the equation. \n\nt^2 = t * t = 1 * 1 = 1\n\nso\nd = 1/2 * 32 * 1\nd = 16 feet
11463
The changes propagate at the speed of light, so I guess the time taken is,\n\nt = r/c where c = speed of light
171383
It takes anywear from 8-9 min! The reason being that the earth does not revolve around the sun in a perfect circle. This answer only counts if I interpreted your question correctly.If not the speed of light itself does not travel.
269338
The speed of light is 3 x 10^10 cm/sec or 3 x 10^8 m/sec or 3 x 10^5 km/sec.\n\nWavelength = speed of light / frequency\n\nBe sure you use the right units.
214684
No massive object can move at the speed of light, so the question is ill-posed. Sorry.\n\nAs a physicist, please let me give you a little advice: don't loose your time with this nonsense, try instead to focus on the mathematics of the relativity theory, that's the way to understand it. The rest is pseudo-science.
759079
The way that I figure it, the shadow is moving at 1500 ft/s after 0.5 seconds of travel.\n\nBasically, the way that I did it was to make a diagram.  At 0.5s the ball has dropped 4 feet from where it started.  If you draw a diagram with a light 50 feet off of the ground and a ball 46 feet off of the ground (because it fell 4 feet in the 0.5s) then you can draw a line connecting the light and the ball and extend that line all the way to the ground.  Now you have a few right triangles to work with.  There is a triangle between the light, the ball and the place that the ball started from and an from the horizontal line connecting the light and the place the ball started from and a diagonal line from the light to where the ball ended.  Call this angle, theta-1.  Now this is the same angle between the line that shows where the shadow is now, and the ground, call this theta-2 (to distinguish them but their value is equal.  The distance that the shadow is from where the ball will land (30 feet from the base of the light pole) is unknown so we will call it x.  \n\nx can be found since tan (theta-1)= tan (theta-2), so then 4/30= 46/x and solve for x.\n\nNow comes the more complicated part.  We don't really need x, we need the rate of change of x with respect to time.  So we need an equation that gives us x with respect to time and then we need to find the derivative of that equation in order to get the velocity at time, t=0.5s.  So we look at the above equation for x and realize that the number 30 doesn't change with time because it is the distance between the pole and where the ball will land.  What will change with time will be the position of the ball.  So instead of 4 in the above equation, we can use the equation for position that they gave us (16t^2).  And instead of 46 (which was the height of the pole (50 feet) minus the position of the ball, which we just defined as 16y^2) we can put 50-16t^2.\n\nOur equation now is (16t^2)/30= (50-16t^2)/ x.  If we solve for x we get x= (93.75/t^2) - 30.  Take the derivative and we get\n x'= -187.5/t^3.  Which makes sense because the closer the shadow gets to the place where the ball will hit, the slower it will go.  Plug in t=0.5s and I get 1500ft/s.\n\nThere might be an easier way, but I don't know it.
823010
As I write this, the question has been posted for 3 hours without an answer, so I'll give it a try.  When you suspend a slack (not taut) string between two points, the string, having weight, will naturally form a parabolic shape between the points.\n\nEarlier I did the math, figuring out how long the string would be if strung across the 8-inch diameter and falling parabolically to a depth of 10 inches in the center.  This is a "length of curve" problem that involves some very messy calculus.\n\nThat's not the only problem.  Having done the math for the longest string -- the one that crosses the diameter -- I realized that the same problem would have to be repeated for each of the chords that will have shorter distances between the endpoints and shorter depths.\n\nAs a result, after reading your question again, I decided to take a new tack, focusing on the "how do you make" part of your question.  There's still some math, but not nearly as much as there would've been otherwise.\n\nTo begin, think of a circular tennis racket with loose strings sinking in the center.  That's the kind of net we're going to design.  For orientation, think of the 8-inch circular rim as the face of a clock. with 12:00, 1:00, ... 11:00 positions.\n\nFor the general shape, long strings will connect the 12:00 and 6:00 positions, falling parabolically to a central depth of 10 inches, and also connect the 3:00 and 9:00 positions, falling to the same depth.\n\nThere can be shorter strings connecting 1:00 and 5:00, 11:00 and 7:00, 10:00 and 2:00, and 8:00 and 4:00, and falling to a shallower depth (less than 10 inches).  These four strings will all be the same length.  Another set of four still-shorter strings will connect 2 and 4, 10 and 8, 11 and 1, and 7 and 5, again all of equal length and falling to a still-shorter depth.\n\nAll strings will be parabolic.  The longer and deeper strings will be in the center, and the shorter and shallower strings (the chords) will be toward the sides.  There will be an odd number of both vertical (e.g., 12-6) and horizontal (e.g., 9-3) strings because there's one central string in each direction and an even number of strings on the sides due to symmetry.\n\nTo form the net, the strings will all be woven as they interlace with each other.  I imagine that your net may be constructed with fishing line.\n\nHaving described the general structure of the net, we now have to do some math to figure out the spacing of the strings.  Going back to the tennis racquet idea, let's suppose we want the strings to be equally spaced <when viewed from above.>\n\nFor an example, suppose the strings are a half-inch apart.  Let the 12-6 line be the y-axis and the 9-3 line be the x-axis.  We want vertical strings at x-values of 0, 0.5, 1.0, 1.5, ... 3.5 (plus and minus) for a total of 15 strings.  There'll also be 15 horizontal strings for a total of 30.\n\nWe've already said that the central strings will attach across diameters (plus and minus 4) and descend to a depth of 10.  In three dimensions, suppose the bottom of the net is at (0,0,0), and the top circle is in the plane z = 10.  The equation of the parabola for the center strings is z = ax^2, passing through (4,10), so 10 = 16a, or a = 5/8, and the final equation in the x-z plane is\n\nz = (5/8) x^2  (Equation 1)\n\nLet's look at the placement of the string at x=3.  In the x-y plane, the circle has radius 4, and by the Pythagorean Theorem, the y-value is sqrt(4^2 - 3^2) = sqrt 7.  This string will be attached at (3, sqrt 7, 10) in 3 dimensions, with the other end at negative sqrt 7 in the y-dimension.  The distance between the attachment points is 2 sqrt 7.\n\nHow far down will this string go?  To get that, we evaluate Equation 1 at x=3:\n\nz = (5/8)(3^2) = 45/8 = 5 5/8 = 5.625\n\nThis means the <height of the netting> at x=3 is z= 5.625 inches.  But the top of the netting is at z=10, so the <drop> is 10 -
175800
To throw a ball 90 mph you are measuring speed not distance. You do not need to reach a certain distance, just a certain rate (distance/time).
13174
Hi Writholeen, the only flaw in JayNick's answer is the accuracy part. As per the National Institute of Standards and Technology, the exact value of Speed of Light in vacuum is 299,792,458 m/s. By exact, I mean there is no uncertainity till the last digit. \n\nTo be precise, what we usually call the "speed of light" is really the speed of light in a vacuum (the absence of matter). In reality, the speed of light depends on the material that light moves through. Thus, for example, light moves slower in glass than in air, and in both cases the speed is less than in a vacuum. However, the density of matter between the stars is sufficiently low that the actual speed of light through most of interstellar space is essentially the speed it would have through a vacuum, so we don't make much error by ignoring the difference.\n\nTo make it easier for you to convert, here is the speed in alternative units :\n\n(1) 983,571,056.4304461 foot/second\n(2) 11,802,852,677.165352 inch/second\n(3) 1,079,252,848.8 kilometer/hour\n(4) 299,792.458 kilometer/second\n(5) 582,749,918.3585312 knot\n(6) 880,991.0899527 mach\n(7) 670,616,629.3843952 mile/hour (mph)\n\nThese values are accurate to the 7th digit of decimal. \n\n\nFlaw in JayNick's answer:\n\nThe speed of light in kilometres per second is 279,600 kilometres/sec. Even if you approximate the actual value , 299,792.458 kilometer/second by round off the last three digit, the value would have been 300,000 kilometers per second.
160198
This is a very interesting question and most people don't understand it.\nTHE value of the speed of light  is "C" in vaccuum IRRESPECTIVE of the refernce frame. Light will travel at velocity C (relative to you) even if you are going at C. \n\nSounds crazy? This is at the fundamental of Einstien's assumption which was later proved right. Common sense is often wrong.
540992
HOW LONG IS A LENGHT OF STRING?
63805
Actually there is no speed of dark.\nBut according to me it should be the same as the speed of light because as soon as the light goes the dark appears.So it must be as the speed of light i.e. 3* 10^8 meters/seconds
219832
The speed of light is indeed a constant.  What's important is that it is a constant to every observer.  If you are traveling toward a light source at 1000 m/s, you will measure the light's velocity at 300,000 m/s.  If you are traveling AWAY from the light source, it will still measure 300,000 m/s.\n\nLight IS the absolute speed limit. (Of this universe anyway)
185055
The equations imply that it is impossible for any object with >0 rest mass to accelerate from below the speed of light up to exactly the speed of light. This is because, as has been mentioned, that would take an infinite amount of energy.\n\nThe equations don't prohibit an object traveling faster than the speed of light from the outset though. Photons have zero rest mass, and the instant they are created they are traveling at exactly c. If it were possible for a particle to exist that was traveling faster than c from the instant it was created, this would not be a problem for relativity. I'm not sure what it would imply about it's mass though. The name for such a particle is a tachyon (although there is no evidence these actually exist as far as I'm aware).
11469
the speed of light = 299 792 458 m / s
418308
The speed of light in a vacuum is exactly 299,792,458 metres per second (or 1,079,252,848.8  km·h-1, which is approximately 186,282.397 miles per second, or 670,616,629.4 miles per hour).