A ball is dropped from a height of 12 ft. Each time it bounces....?
955166
D = d0 + d sum(%b^N) for N = 1 to 5 bounces; where d = 12 ft. the initial height and %b = .6 and d0 = (12 or 24) depending on if raising the ball to height d = 12 is included in the total distance.\n\nYou can show this by:\n\nN = 0 the starting point with d0 = (12 or 24 ft.), depending on if you count raising the ball to 12 ft. as well as dropping it to start the bouncing.\nN = 1 the first bounce; so (.6)^1 = 60%; so the distance it travels here (d1) is d (%b) up AND also d (%b) down to start the next bounce.  Thus, d1 = 2 d (%b^1)\nN = 2 the second bounce; so  (.6)*(.6) or 60% of the 60% first bounce; so the distance d2 = 2 d (%b)^2\nN = 3 the third bounce distance would be similar; so d3 = 2 d (%b^3) and so on up to N = 5.\n\nIn which case, D = d0 + d1 + d2 + d3...+ d5 = d0 + 2 *12(.6 + .6^2 + .6^3 + .6^4 + .6^5) = d0 + 2 * 12 sum(.6^N) for N = 1 to 5; If you count raising the ball to the initial height, then d0 = 24 = 2 d; otherwise d0 = 12 ft., the distance of the first drop.\n\nYou can do the arithmetic.
896561
If the object starts out at a height h0 and is dropped, then all the potential energy is converted in kinetic energy\n\nm*g*h0 = 0.5*m*v0^2\n\nso that at impact the object is moving at a speed v0\n\nv0 = sqrt(2*g*h0)\n\nAfter bouncing off the surface the object is moving with a speed v1\n\nv1 = e*v0\n\nwhere e is the coefficient of restitution. In a purely elastic collision e=1, but in the real world e<1. Using the conservation of energy again\n\n0.5*m*v1^2 = m*g*h1\n\nwhere h1 is the new height to which the object bounces.\n\nh1 = 0.5*e^2*v0^2/g\n\nAt moment of next impact the object is moving at speed v1, but rebounds at speed v2\n\nv2 = e*v1 = e^2*v0\n\nThe height h2 is then\n\nh2 = 0.5*e^2*v1^2/g = 0.5*e^4*v0^2/g\n\nThis continues for however many bounces you wish. Recall now that v0^2 = 2*g*h0 so that the heights are\n\nh0 = e^0*h0 (shown only for consistency with the following)\nh1 = e^2*h0\nh2 = e^4*h0\nh3 = e^6*h0\n...\nhn = e^(2n)*h0
711024
Look, task 1,3,4 are very alike. We shall use the same formulas there:\n1. let us find energy of the body at A, at B, and at C\nat A:   mV^2/2 + mgH = m(V^2/2 + gH)\nat B H=0, so the energy constitutes the kinetic energy mV(b)^2/2\nat C: mV(c)^2/2 +mgH(c) = m(V(c)^2/2 + gH(c))\nthe energy does not change (not dissipates), so energy in A equals the energy in C:\n\nV^2/2 + gH = V(c)^2/2 + gH(C), so \nV(c) = sq.root {V(a)^2 + 2g(H(a) - H(c))} = \n = sq.root{64 - 2x2x9.81} = 4.976 m/s\n2. The force of gravity for the mass 100 kg will be \nF = mg = 100kg x 9.81 m/s^2 = 981N\nAs the mass is moved along steep surface, some part of the gravity force will hinder the movement. We can fin the component of the gravity force along the surface using the triangular 20m and 5 m proportion, (because the triangles are similar), so\nFg(along) = 981N x 5/20 = 981N/4 = 245,25N\nWe have to apply the force 300N to move the body along the surface, hence we overcome the gravity component 245.25, and the rest is for friction.  \nSo the Ffriction = F - Fg(along) = 300N - 245.25N = 54.75N\n\n3. The formulars are as in problem 1. \nat the start   Es = Ekinetic + mgH\nat the highest (final) point of the bounced ball\nEf = Ekineticf + mgHf,\nHere we have to make some remarks (very important)\nIf the question would be "How high the ball can bounce",\nthen we could assume that the highest point would be when the ball bounces vertically up, and at its highest point it has zero velocity, no vertical one, no horizontal one. Than we can makethe energies equal (assuming no dissipation) and find Hf:\nmV^2/2 + mgH = mgHf\nso,  Hf = H + V^2/2g = 4(m) + 36(m^2/s^2)/2x9.81(m/s^2) = \n             = 4(m) + 1.835(m) = 5.835 (m)\nIn case the ball bounces not vertically, but along the steep line, at his highest point it will have zero velocity too, so the answer will apply to the case. \n\n4. The velocity in B will be found using the formula for energy at starting piont and at B:\nmV^2/2 + mgH = mV(b)^2/2,\n\nV(b) = sq.root{V^2 + 2gH} = sq.root{16 + 12x9.81} = 11.56 (m/s)
526447
I actually had my class conduct an experiment on this very thing.  The problem was: does the percentage of energy lost from a ball bouncing increase, decrease, or stay the same as the height increases.  This is akin to your question, because the height at which it bounces back to depends on how much energy is left after the bounce.  \n\nWe found that the percantage of energy lost stays a constant, which means the the answer to your question is yes.\n\nFor example:\nIf you start with a height of 10 meters and drop a 1.0 kg ball, it will have 98 Joules of potential energy (PE = mgh).  If it loses 10% of its energy on the bounce, it will have 88.2 Joules of kinetic energy, which will translate to a height of 9.0 meters.  \n\nIf you double the original height, you will have 196 Joules of energy.  After losing 10% from the bounce, you are left with 176.4 Joules, which translates to a height of 18 meters after the bounce.
495581
a round ball I guess LoL
930941
12, same as a classical soccer ball. \nIt is a dodecahedron whose edges are capped with equilateral hexagons.
840035
Ok, we're going to assume the 4.6268 cm is in addition to the 58.8cm, so the ball doesn't stop until it has fallen 63.4268 cm.  Therefore it has lost .0833*.634268*9.8 = .51778 Nm of energy.\n\nSince all this energy goes into compressing the spring, and its energy is .5kx^2, k = 2*.51778/.046268^2 = 483.74 N/m
227905
When you lift a ball off the floor, you transfer energy to it. This energy is stored in the gravitational force between the ball and the earth and is called gravitational potential energy. When you release the ball, its weight makes it accelerate downward and its gravitational potential energy gradually becomes kinetic energy, the energy of motion. When the ball hits the floor, both the ball's bottom surface and the floor's upper surface begin to distort and the ball's kinetic energy becomes elastic potential energy in these two distorted surfaces. The ball accelerates upward during this process and eventually comes to a complete stop. When it does, most of the energy that was initially gravitational potential energy and later kinetic energy has become elastic potential energy in the surfaces. However, some of the original energy has been converted into thermal energy by internal frictional forces in the ball and floor. The distorted ball and floor then push apart and the ball rebounds into the air. Some or most of the elastic potential energy becomes kinetic energy in the ball, and the rising ball then converts this kinetic energy into gravitational potential energy. But the ball doesn't reach its original height because some of its original gravitational potential energy has been converted into thermal energy during the bounce.
299848
You only need to change the temperature of the ball, since the floor and air temperature will only have a very minor effect. I'd suggest immersing the ball in water of various temperatures, stirring occasionally, leaving it in until equilibrium is assured (5 or 10 minutes, maybe). More water will mean less change toward room temperature while heating/cooling the ball. Measure the water temperature, then remove the ball and perform the bounce as soon as possible afterward. If you are testing temperatures below freezing or above boiling you have to allow more time for the ball to reach the temperature and take the temperature of the freezer or oven. \nHow to catch and measure the bounce height is tricky. It would be best to have a background (including the measuring scale) that visually contrasts with the ball, and it would be helpful to have an assistant who drops the ball while you maintain a steady viewing position. Of course the drop should be done with no added energy (no upward or downward velocity). Before starting the experiment, try different ways to catch the bounce height. Your eye should be level with the bounce height, which of course is only approximately known before the drop. Bounce height should be measured at the bottom of the ball, or at the top with the diameter subtracted later. You can make a good guess at the height (colder=less bounce, warmer=more). You may have to make more than one drop at a given temperature if the bounce height is too far off eye level.
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.
648817
a) Supposing that s0 = 0. We  have v0 = 32ft/sec\n=> s = -16t^2 + 32 t\n\nb)When t = 1, s=-16 + 32 = 16 ft above the ground\n\nc)When s = 0 again:\n0 = -16t^2 + 32 t\n=>t(-16t + 32) = 0\nThe equation is equal to zero when t=0 (the time the ball was thrown) and when t = 2 => The ball will hit the ground at t=2\n\nd)ds/dt = -32t + 32\nThis is equal to 0 when t=1\nat this point s = 16.
498095
A two piece ball like Pinnacle will go the distance.  Balls that spin will not go as far
52009
<shake><shake>\nMagic 8 Ball Says:\n\nAsk Again....\n\n\n<shake real hard several times>\nMagic 8 Ball Says:\n\nYes!...
239426
It's a basketball hoop..  which hangs 10 feet above the floor.
462053
it is a sphere while in suspension and then the classic teardrop shape while falling.
608626
Due to law of conservation of energy. The kinetic energy it gains during falling is restored in the rubber at the time of collision while in other balls it is lost in sound & heat. So the crazy ball bounce much more than other balls.
470687
At the top of the halfpipe, the ball is just sitting there with potential energy stored up (PE = mgh, where m = mass, g = acceleartion due to gravity, and h = height).  As the ball rolls downward, its PE lessens and its kinetic energy increases (KE = 1/2(mv^2), where v = velocity of the ball).  At the bottom of the halfpipe, all PE is gone and KE is at maximum.  As the ball goes up the other side, the KE decreases and the PE increases.  In a perfectly frictionless halfpipe, the ball would return to the same height as it was before it started rolling, and the KE would go to 0 and the PE would be at its maximum.\n\nIf you just dropped the ball (no halfpipe - just a freefall), the KE just before impact would be the same as the PE before dropping.  The two are inversely proportional - as one goes down, the other goes up.\n\nNow if you knew the exact shape of the halfpipe, you could start plugging in numbers to find the PE and KE of the ball at various points.