A man of mass 65kg is on a frictionless surface he throws a ball of mass .350kg with a velocity of 28m/s.....?
924429
The ball\nVt= Vo + at\nVo= 0 hence Vt= at\nF = m * a = 0.350 * a\na= F/0.350\n28 = F/0.350 * t\nF= 28 * 0.350 /t  = 9.8 /t\n\nThe man\nt man = t ball\nF man = F ball ( Newton's law)\nF = 9.8/t\nF = m * a = 65 * a\n9.8/t = 65 * a\na = 9.8/(65*t)\nV=a*t\nV= 9.8/(65*t) * t = 9.8/65 = 0.151 m/sec
765131
#1 use conservation of energy:\n\nThe work done by the force of the raquet is equivalent to the kinetic energy of the ball.\n\nF*d=1/2*m*v^2\nyou have all of the variables except F.\n\nF=.5*.56*30*30/.5\n=504 N\n\n\n#2\nIt will take a force of:\n50*9.8*.5\n=245N\nto overcome the static friction\n\na) The net force accelerating the box is:\n250-50*9.8*.4\n=54\nwhich =m*a\nso a=54/50\n=1.08 m/s^2\n\nb) If the force is pushing downward it is an insufficient force to overcome static friction or keep the box moving against the kinetic friction.  The box is at rest\nIf the force is pushing upward at a 45 degree angle, then the force of static friction is\n(50*9.8-sqrt(2)*250)*.5\n=156N\nand the hoizontal component of the force is\nSqrt(2)*250=176\nwhich is strong enough to overcome the static friction.\nSo acceleration again using\nF=m*a=176-(156*.4/.5)\na=51.2/50\n=1.024 m/s^2\n\n\nj
200510
The answer given by "Amanda" is actually quite funny, I like it.\n\nWork = Force * Distance applied\nIf a constant force of 45 Newtons is applies over a distance of 4 meters, the work done would be 180 Joules.\nI hope we can assume this 45 Newton force is being applied parallel to the ramp due to the weight (force of gravity) of the object.
167187
The person opposite of the side which the ball bounced twice.
143729
Final veloc. when he reaches the land:\nV=sqr(2*g*h)= 8.743 m/s\nKinetic Energy= Ec= 0.5 * m* V2 = 1872,780 \nThis turns into potential Energy Ep= 1872,780  = mgh\nHe stops in h=0.8m , hence: \nForce = m g = 1872,78 / 0,8 = 2340,795 Newtons\n\nHe has two knees, so F= 2340, 795 / 2 = 1170,488 Newtons in each knee.
750261
His balls were dribbling
69030
Initially, the ball is pushed, it slides. There is no rotation, as stated in the problem/question. This means there is no rotation at all. That all the velocity is translational.\n\nFor this scenario think of the ball as a solid block that has been pushed and it sliding across the floor.\n\nAs stared, at some time (t1) the ball begins to roll.\n\nAt this time, the ball's motion is both rotational as well as translational. In other words, it rolls as well as moving forward. This motion is what takes place at the *contact point* of the ball to the floor.\n\nIf the ball rolls, it is not slipping.\n\nIndeed, the friction causes the linear motion to decrease, and eventually stop the ball.\n\nThe rotational motion of the ball is also undergoing friction. A rolling sphere's motion is very subtle and often misunderstood. By virtue of the ball's mass there is a very tiny (tiny is relative to the mass and material) flat spot between the ball and the surface upon which it rolls, whereby a small torque gives rise to a resistance of the motion. This is why a rolling sphere slows. But.....since there is friction, the velocity of translation, is related to that of rotation by the following:\n\nv^2 = 2gh/(1 + I/Mr^2)\n\nWhere M is the mass of the ball and "r" its radius, I its moment of inertia. This is common notation. If there is no rotation, then the moment of inertia term drops out (I = 0) and you are left with a velocity which is strictly that of a translating body......kind of like free-fall.\n\nIf the ball is given one initial push to give it the initial velocity Vo, then unless there is some other forcing function at work, friction will act to decrease the motion of the ball, and all said velocities thereof. None of the velocities will increase.\n\nIf you know of a situation where such a case can exist where the ball increases velocity without doing anything to it.....make sure you mention my name when you pick up the your Nobel Prize.\n\n:)
1010100
ok, so if we're assuming that it was a head on elastic collision, and all the energy was transferred then there are two possible situations: if the 25lb ball was sliding then the total energy transferred will be equal to the amount it had originally (since we assumed elastic collision) that is equal to 1/2mv^2...where mass would usually be measured using the metric system. \nif the ball was rolling then it has not only translational energy (1/2mv^2) but also rotational energy which is equal to 1/2Iw^2 where I = inertia of a ball...if it is a solid sphere the moment of inertia is equal to 2/5mr^2 (m = mass, r = radius) and w (omega) = rotational speed in radians per second...which is equal to velocity/radius. so the amount of energy transferred would be equal to 1/2Iw^2 + 1/2mv^2.
461302
So far, you have a starting position and an ending position.  If this were a physics question, you'd also need numbers like the mass of the ball and the initial launch velocity.\n\nIn general the equation of motion is given by\n\ns = s0 + v0t + at^2\n\nwhere you need or have the final position (which we have), initial position (which we have), initial speed (which we're looking for, gravity (a known constant), and the time in flight (which we can find using other equations).  Generally, things will go up and down parabolically, not in a straight line, so we need more information about this to answer your question.\n\nNow if you're just looking at a straight line path, then you can use some trig to find at least the angle to the ball (you can't find the speed if you're assuming straight line travel).  In this case, if the person is 1.7 m, and he is standing 5.8 m from the basket that is 3.05 m away, you have a right triangle where the horizontal leg is 5.8 m and the vertical leg is 4.1 m (the difference between the height of the basket and the player, assuming that the ball is thrown from 1.7 m above the ground).  Using the tangent function, you can find that angle:\n\ntan x = opposite side / adjacent side\n= 4.1 / 5.8\n= 0.7069\nx = 35.26 degrees\n\nBut again, laws of motion dictate that the ball follow a parabola to the basket.  But at least we can say that the launch angle has to be at least 35.26 degrees.
1106414
puting the ball threw someones legs
119016
i guess it would depend on how much he practices, i dont believe there is a set distance that someone should be able to throw. just have him practice enough to get better, but not too much to injure himself
857410
Actually, the wall having no K.E. is okay, since the speed of the ball is the same before and after the collision. Momentum IS transfered to the wall and the earth supporting it but the mass of the wall (and earth) is so great that the corresponding change in velocity is undetectable...
792153
‘impulse’ or momentum\n=m*(v2-v1)=0.42*(-17-4.1)=-8.862 kg*m/s
680534
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
494819
MORE OR LESS 135 KM/H.
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).
484307
in air= 340m/s\nin water=1500m/s\nin solid=4500m/s