How do I measure the velocity of a falling ball using a strobe light?
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You need a camera, an accurately timed strobe light, and a background with a measured grid. In a darkened room, you open the shutter of the camera, turn on the strobe and drop the ball. You will end up with a picture with multiple images of the ball as it fell. Using the grid on the photo, you can determine the distance it fell between flashes. Knowing the time interval between flashes, you can compute the average velocity of the ball during each time segment.
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An animation flash demo with stepwise calculation is given\nhttp://www.technocs.vipzone.biz
1168507
Here is one example...the bouncing ball. Just google that in and you get a bunch of results.
341507
You mean find "g" as if you didn't know the constant? Okay.\n\nPretty simply really. Have the height measured that you're going to drop it from to where it's going to land (this will be your "h"). Make it a high one (one meter is enough). Measure the mass of the object (that's the "w"). Then when you do the experiment, time it. Start it at the instant you let it drop and stop it once it touches where ever it lands. That's enough information to find "g." Just solve for "g" then that would be the answer.\n\nActually, for an experiment, you should solve for "g" as a formula before you start doing anything else. So the formula would be like this:\n\ng = (2h)/[(t^2)(w)]\n\n(t^2 is "t squared")\n\nSo just have that formula then do the experiment: measure height of the drop, the mass of the object, then time it when you drop it (you need good reflexes when you drop it, as it falls fast).\n\nPS Penguins!!!
193098
There are two variables you'll need to measure experimentally.\n\n1)The height of the ball just before decent\n2)The max height of the ball when it bounces back.\n\nIf you dived 2) by 1) you'll get a fraction that is constant idependently of the heighth dropped (assuming you don't drop it to high!!), hence, its called a 'constant.'
433903
The ball will reach a maximum height, then the velocity of gravity (9.8 meters/second) would cause it to travel downward until another force acts upon it (according to Newton's laws of motion). However, this does not account for other forces, such as air resistance. The rate at which the ball falls and the air resistance acting upon it will depend on the properties of the ball (surface area, weight, etc.). If the ball is inside a vacuum, it is said to be in free fall (as the only force acting upon it is gravity) and there is no other force acting upon it. If the ball is allowed to fall long enough, it will reach a point of terminal velocity, where it cannot fall any faster (the force of air resistance will equal the weight of the object)
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Don't even worry about the rock yet. The ball takes a certain time to drop 45m vertically and in that time it goes 30m horizontally. Air resistance is negligible so the horizontal velocity is constant. In the vertical direction, the ball starts motionless and accelerates and 9.81 m/s^2.\n\nTime to fall 45m\n45 = (1/2)(9.81)t^2\nt = 3.0289s\n\nhorizontal velocity:\n30 = 3.0289*v\nv = 9.9m/s
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For a back row spike I found 26.7-27.6m/s (mean).  Sorry, but that was all I could find!  I was going to do a study on it - but then I switched out of biomechanics into exercise physiology.  You could do it by calculating time elapsed between contacting the ball and the ball reaching the floor.  You would have to know the height at which the ball was contacted and how far it traveled horizontally from the point of contact.  They you find the hypotenuse using Pythagorean theorem and divide this number by the time to get an estimated velocity.
710256
You got the CG just right.  Your method might work with enough concentration., but I would look at making the angular momentum come to zero:\nL = Iω = (2.7*.325²  + 1.3*.675²) * 26rpm*2π/60 = 2.389 kg m²/s\n\nTo stop the movement the angular impulse must equal the - of the angular momentum:\n\nT * t = L→ T = L/t = 2.389/5.8 = .4118 Nm (ccw)
447491
The whole thing in pool is that the angle of inflection is equal to the angle of reflection.  I wonder, I've never tried this, if using a laser pointer to show the line of sight and the place on the ball where the cue ball should hit.  You could also use a mirror and a laser to find the position on a the bumper to shoot the cue ball for a bank shot.  If you move the mirror along the bank with a laser behind the cue ball pointing at the mirror, the only time it will shine on the ball intended to be struck will be at the position you would like to aim the cue ball in.  \n\nI just made all this up, so I don't know if it would work.  But, it would be a good visual way to describe what you are trying to explain to your friend.
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1)Start by modelling the speed of a ball in one direction; take into account friction force (negative drag) that is equivalent to negative acceleration. Loop until speed is equal or less than to zero. Then, stop.\n\n2) Do Exactly the same, now in a 2-D space: model how'd the would move given an initial vector speed.\n\n3) Model a "Hit" on a ball: given an instant (vector) speed plus a "Hit Vector", compute the resultant vector; then use that vector as the initial speed again.\n\n4) develop a "Hit?" function that returns true if the difference between the centers of the balls is less than the diameter, or if the center of the ball is closer than a radius to a wall (the stick can be thought of as another ball)\n5) Every time Hit? is true, recompute vector speed for every involved ball.\nVoila! you have your game. Let me know if it's useful and the results.
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1. You need to know the initial height of the ball from the ground.  I will assume from ground level.\n\n2. Calculate the vertical and horizonal components of the speed.\na^2 + b^2 = c^2 where c = 80 and since it is a 45 angle a=b\n2a^2 = 80^2\na = 56.6 mph (both vertical and horizontal speeds are the same)\n\n2. Now take the vertical speed (56.6mph) and find time it takes to hit the ground.\nSince acceleration due to gravity is -32.2 ft/(s^2)\n56.6 mi/h * 1/60 h/min * 1/60 min/s * 5280 ft/mi = 83 ft/s\nNow using\na=(v2-v1)/t where a is acceleration, v1 = inital velocity, and v2 = final velocity, and t = time it took\n-32.2 = (0 - 83)/t\nt = -83 / -32.2 = 2.58sec\nsince that is the time for the ball to finally reach the top of the arc, multiply by two to get the complete arc\ntotal time in the air = 5.16sec\n\n3. Calculate distance travelled in that time.\nI am leaving air resistance out due to incomplete data (you need mass of the ball, resistance will also change due to the shape, the barometric pressure, etc.)\n5.16s * 83 ft/s = 428.3 ft travelled\n\nEdit:\nSince the distance from the ground is small compared to the velocity of it falling, we can estimate just an additiona 0.04 secs so total time is now 5.20 secs\n\n4. Drag equation\nD = Cd * A * 0.5 * r * V^2\nCd = Drag coefficient\nA = reference area (half surface area of baseball 2pi * r^2 = 3.53 ft^2)\nr = density of air (20 C and sea level is 1.2 kg/m3)\nr = 1.2 kg/m^3 * 2.2 lb/kg * 0.0283 m^3/ft^3 = 0.0747 lb/ft^3\n\nCd can't be calculated since it is different for every object and situation.  Only way for Scientist to find it is using a wind tunnel.\nI cheated and looked it up for a standard baseball it is 0.007 lb/ft\nD = 0.007 * 3.53 * 0.5 * 0.0747 * V^2\nD = 0.0009229 * V^2\nDi = 0.0009229 * 83^2 initial force from drag \nDi = 6.36\nuse F = ma to get the negative initial acceleration\n6.36 = 0.3125 * ai\nai = 20.4\n\n-20.4 = (d/5.20 - 83)/5.20\nd = 120 ft\nThis is only true if the drag stays constant, but it decreases as the velocity decrease (exponentially).\n\nAn estimation can be done by calcing the acceleration after each second, then decreasing velocity due to that acceleration.\nV(i) = 83.0, Drag(i) = 6.36, a(i) = -20.35\nV(1) = 62.7, Drag(i) = 3.62, a(i) = -11.59\nV(2) = 51.1, Drag(i) = 3.41, a(i) = -7.70\nV(3) = 43.4, Drag(i) = 1.74, a(i) = -5.55\nV(4) = 37.8, Drag(i) = 1.32, a(i) = -4.22\nV(f) = 33.6\nAfter adding up the distances travelled after each second (including the 0.2 sec at the final velocity), I come up with an estimate of 284.6 ft travelled.  \n\nThis is still a desent estimate, but there is still additional factors to consider.  The ball may reach terminal velocity on its way down which will increase the time returning to the ground.  This problem at first glance looks simple, but as you can see, it gets complex very quickly.
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sorry, but dont know what Laser Diode is ??\nLED is Light Emitting Diode.\n\nYou need a transmitter and receiver when the diode emitts the lights it bounces back from the object and the receiver put at same point as transmitter will record the time it took to come back. so the time to the object is half of that time\n\nnow distance D = V (velocity here speed of light) * t (time)\n\nthis mathod is not easy to construct and since speed of light is very high, this device doesnt work for small distances\n\nthanks\nSANDIP
788507
If you have a video camera and good editing software (iMovie), you can measure the speed of sound. Have a friend set off a firecracker a long distance away, and tape it. Then count the number of frames between the flash and the bang.
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Get a softball and a flashlight.\n\nHold the softball in front of you and hold the flashlight beside it so it's beam shines onto the ball.\n\nIf the light is coming from beside it it's like a half moon.\n\nMove the flashlight so that it's shining on the ball but from slightly further away from you and the ball.\n\nIf the light is coming from slightly behind and beside the ball it'll be like the waning moon.\n\nPut it another way, the moon has day and night on it's surface as well. The bright part you see is the half that's being illuminated by the sun.\n\nOn the moon the day night cycle takes around thirty days. The night is fifteen earth days long and the day is fifteen earth days long.
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FOLLOW THE BOUNCING BALL
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1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj