if you drop a bowling ball and a basketball from the roof, which will hit the ground first?
81552
If you disregard air resistance, both land at the same time.\n\nThe Earth's gravitational pull affects everything equally.
312163
the first after collision moves in an opposite direction and the second ball moves towards the direction of collision\nfirst ball moves with a velocity of  16.66cm/s\nsecond ball with 66.66cm/s
963941
Bound's hubby here:\n\nFirst off, how certain are you that the bullets are penetrating the bowling ball and not ricocheting off the round surface of the bowling ball? Second, you should be aware that the core of the bowling ball is almost solid cork, which will absorb more shock than you can dish out. Your bowling ball is really a poor choice in target from a safety stand point.\n\nGood luck.
859137
first thing u should do is do draw the path of the projectile,  using only imagination, the ball will definetly fall down as it moves away from the table edge until it finally hits the floor.\n\nprojectile path is always parabolic, and the direction of the velocity vector is tangent to the path of the projectile at any point.\n\nat the initial state, the velocity vector was pointing left or right according to your assumption, but as the parabolic path falls down, the velocity vector points down with it, when the ball hits the floor the velocity vector will be perpendicular to the horizontal in the downward direction.\n\nso at 45 degrees downward, the velocity vector has two components, on in the horizontal V*cos(theta) and another downward V*sin(theta), to get how high the ball at this instant apply the same laws but use V*sin(theta) instead of V ,  and to find how far from the table use V*cos(theta) instead of V.\n\nhope this helped.
879264
'Large' and 'small' probably refer to the masses of the balls, rather than the sizes. Assume one ball (with mass m1) is initially directly above the other (mass m2), and both are dropped at the same time from rest (with a slight vertical space between the two).\n\nThe lower ball hits first, bounces up and collides with the upper ball near the ground. The two  have about the same speed v = sqrt(2gH) going into this collision because both balls have fallen from approximately the same height. If the two masses were equal, then both would simply reverse direction, the top ball going upward at speed v and the bottom ball going down (for the second time) and immediately up at the same speed v --> both balls would continue up at the same speeds to the original height H.\n\nNow suppose the masses are slightly different with m2 > m1. Then the initial and final velocities (taking the upward direction as positive) for the collision are respectively:\n\nv1 = -v\nv2 = v\n\nand:\n\nv1 = v + x1\nv2 = -v + x2 \n\nwhere x1,x2 << v. Energy and momentum conservation require that:\n\n(m2-m1)v = m1(v + x1) + m2(-v + x2)\n(m1+m2)v^2 = m1(v + x1)^2 + m2(-v + x2)^2 \n= m1(v^2 + 2vx1) + m2(v^2 - 2vx2) (Approximately)\n-->\n2(m2-m1)v = m1x1 + m2x2\n0 = 2m1vx1 - 2m2vx2\n-->\nx1 = (m2/m1 - 1)v\nx2 = (1 - m1/m2)v\n\nSo the top (lighter) ball ends up with a higher speed (m2/m1)v and the lower (heavier) ball ends up with the lower speed (m1/m2)v. This is reflected by the one bouncing higher than H and the other lower than H.\n \nThere can't be more than one collision since the top ball comes out of the first (and only) collision with a higher speed than the bottom ball so the latter can never catch up to it. If the mass ratio were reversed (heavier ball on top) the situation would be much more complicated since there could be multiple collisions.
14021
The ball would bounce higher if the container kept a vacuum in it the entire time.  The first thing to understand is that there will not be pressure on top of the ball, there will be pretty much equal pressure around the ball from all sides; the air will spread evenly throught the container very rapidly.\n\nThis air will create drag, which will slow the descent of the ball and reduce the kinetic energy and momentum of the ball.  Drag and friction are called non-conservative forces, because when kinteic energy is lost to them, it is not transferred to another object, rather it is gone and not retrievable.  (It actually does get transferred, because no energy can truly be destroyed, but it is not retrieable.)\n\nSo what we have so far is a ball that is falling through dense air (greater pressure of a gas means a greater density of the gas), which is putting drag on the ball.  This means the ball does not fall as fast as it would in a vacuum, and the faster a rubber ball falls.  We then suck all the air out instantly, and the ball will bounce back up unrestricted by drag on its way up, but it would have bounced higher if it was not slwoed down by air during its fall and had a higher impact speed.
705043
As the ball hits the ground, its momentum causes its elastic body to compress itself against the ground until enough force had been applied through enough distance to cancel the momentum of the ball and bring it to a stop.  At the instant the ball comes to a stop, it has energy stored in it due to the compression of its body against the ground.  This compressive energy is then released in pressing against the ground, causing the ball to accelerate upward.
434934
I am not sure I understand your question, but here goes. If something is higher it has more potential energy than if it is lower. If you drop something from 10 stories high it will get damaged. If you drop something from only and inch it will probably be all right. It also matters how big the object is so PE (potential Energy)=m(mass)xh(height). Hope this helps.
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.
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)
728879
basketball. When you shoot the ball to make a basket, the force with which it is thrown, combats with gravity. This determines at what point the ball will start to fall towards the basket.
200232
depends on the kind of fall and the trajectory
436294
FOLLOW THE BOUNCING BALL
744710
Well from the right trajectory it Will flip over about 12345678778653789875653654567709 times, followed by 1300000 barrel rolls and land on the Taj Mahal. Immediately it will loose its own balance and fall right on to a car roof with a mattress on it. That will carry it to a pie factory where it will land in a pie with its kidney first.
604227
a fork ball goes down and go etiher straight, right, or left and a sinker just goes down
579002
First I'll assume this isn't a trick question, i.e., that the balls start with the same speed and have equal densities. They are rigid and roll on a rigid surface. What slows them down? Air resistance and going uphill. Not friction because they are rolling, not sliding. \nWith the same initial velocity, they have the same translational momentum per unit mass. However, another factor is the angular momentum due to rolling. Let ball 1 have a radius r=1 and a mass m=1, and ball 2 have r=2 and consequently m=8. If each ball has an initial velocity of 1, ball 1's angular rate w=1 rad/sec and ball 2's w=0.5 rad/sec. The moment of inertia I of a solid sphere = k*m*r^2 (k=0.4), or I=k for ball 1 and I=32*k for ball 2. The angular momentum = I*w, thus I=k for ball 1 and I=16*k for ball 2. While both balls have the same translational momentum per unit mass, ball 2 has twice the angular momentum per unit mass and therefore a greater total momentum per unit mass, which means a greater force * time per unit mass is needed to stop it. Going uphill provides the same force per unit mass on each ball, so ball 2 goes further uphill. Air resistance per unit mass is greater for ball 1 since air resistance depends on area, and ball 1's area/mass ratio is 1 vs. ball 2's ratio of 0.5. Again, ball 2 goes farther.\nEDIT: The balls have different momentum per unit mass but the same energy per unit mass, so my argument above about uphill travel is wrong; they go the same distance uphill. But air resistance still favors the larger ball. So answer 5 is the best answer so far. Another real-life factor would be: suppose the balls are on a rough surface or face an effective drag of a surface such as a rug. A larger ball wins on both. The larger radius means smaller vertical motions for a given scale of roughness (e.g., gravel size), so less energy wasted per unit mass. And you can visualize how a deep-pile rug would stop a rolling golf ball much faster than a bowling ball.
230718
Answer:\n8 seconds\n\nProof:\nThe problem is really a question of finding what value(s) of x make y equal to zero in the given equation.\nWhen y (the height or distance) equals 0 then the ball is either leaving the ground or hitting the ground. The values of x (the time) that make the height equal to zero tell us when this is happening.\nThe given equation is an inverted parabola so there may be two points where y = 0.\n\n1) Solution 1: x = 0\ny = -16x^2 + 128x\ny = -16*(0^2) + 128*0\ny = -16*0 + 0\ny = 0 + 0\ny = 0\nBut x=0 refers to the time when the ball is first thrown. We want to find the time when the ball lands.\n\n1) Solution 2: x = 8\ny = -16x^2 + 128x\ny = -16(8^2) + 128*(8)\ny = -16*64 + 1,024\ny = -1024 + 1,024\ny = 0\nWhen x (the time) equals 8 seconds, y (the height) equals 0. So ball hits the ground after 8 seconds.\n\nFor your information, after 4 seconds the ball reaches its maximum height of 256 feet.