a ball is thrown vertically upward.?
912781
Its velocity will be zero when it reaches its maximum altitude.
495603
Assuming that the energy of the ball remains constant (i.e. no air resistance), the speed will be the same. The VELOCITY, however, will be -35 m/s.\nSpeed and velocity are NOT the same!
103128
When you toss the ball the height you toss it is up to you. I tell my players that their serve should be in a rhythm. Start counting from when you start your motion of your toss. One count is from standing still to dropping your tossing hand down w/ the ball. The upward movement of your hand is the two count and the three count is the release of the ball untill your contact it with your serving hand. The toss is done in front of your body. Keep it front of your body so when you step for your striking motion you will be almost under the ball. When making contact with the ball, the ball should be slightly in front of your body, your arm extended, elbow elevated. Contact the ball with a open, slightly cupped, firm hand. Upon contact, snap your wrist giving the ball a downward spin.
642743
think that the ball is part of ur foot  and throw your leg where you want ur foot to go. Also build up your muscle in ur legs so when you hit the ball true it will go straight. And get a decent ball and boots
788255
use conservation of energy:\n\nThe 810 N force did work for 2 m\nthe result is potential energy gained in height working against gravity including the 2m that the 810N force acted on the ball\n\n810*2=7*9.8*(h+2)\nh=1620/(7*9.8)-2\n=21.6m\n\nj
30231
Well when you are pushing a ball your hands feet whatever you are pushing it with sends pressure onto the other objec in this case the ball making it move where we push it
824387
The ball goes from 0 to 9/s in .15 secs\n\nSo:    9meters           1\n        ------------  x    ------                  =9 /.15 m/sec2\n           second          .15secs\n\n                                                      =60 m/sec 2
201479
When you throw a ball in the air it gradually gets slower until it starts falling back towards the Earth.  The faster you throw it the higher it goes.\n\nImagine now if you throw the ball fast enough that the Earth's gravity will not be able to slow it down enough before it escapes into deep space (outside our solar system).\n\nThis speed is the escape speed or escape velocity.\n\n---------------------------------------------------------\n\nMore accurately when you through a ball in the air you give it a certain amount of kinetic energy, as it gets higher this kinetic energy is being converted into gravitational potential energy (gravitiational potential energy is greater the higher you are).  The escape velocity is the velocity at which it has enough kinetic energy that you would have to take it an infinite distance away from Earth before it was all converted into gravitational potential energy.
503790
Gravity always accelerates the ball in the "Y" direction such that the force pulls the ball downward.\n\nEven if the ball stops in midair, it is still being accelerated at 9.8m/sec^2\n\nThis means no matter how hard you hit it in the x-direction, it will always be accelerating at 9.8m/sec^2 in the y-direction.\n\nThat is a major concept in kinematics.  The motion in the x and y directions are independent of one another!
553125
There are two ways.\n\n1. This is the most likely way.  On a day with no wind, throw the ball straight up.  This is quite unlikely and requires a lot of skills.\n\n2. This one is impossible.  Throw the ball so hard that it goes around the Earth once before coming down.  Humans just don't have the power to do this.
664646
just make it really tall, cause the ball goes pretty high but then again it cant do any real damage to it.
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
325128
Ball starts in the middle(generally theres a drop-in hole on the side). Obviously you try to kick it into the other goal.
838111
mgh = 1/2 mv2\ngravitational potential at top of swing = kinetic at bottom of swing\nFind the vertical component that the ball is raised with the 29 degrees and the 2m radius, plug into equation, solve for v. Masses will cancel out. Since at the lowest point, all the gravitational potential is converted into kinetic energy.
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)
714514
a) they give you everything you need.  v0 = 32, s0 = 0, so\n\ns(t) = -16t² + 32t\n\nb)\n\ns(1) = -16 + 32 = 16\n\nc)  The ball hits the ground when s(t) = 0 (the second time).\n\n0 = -16t² + 32t = -16t(t-2) so t=0 or t-2 = 0, t=2.\n\nSo the ball hits the ground when t=2.\n\nd)  The maximum height is attained at the vertex of the parabola.\n\ny - 16 = -16(t² - 2t + 1)\ny - 16 = -16(t - 1)²\n\nSo the vertex is (1, 16), that is, at time t = 1, when the height of the ball is 16.
658395
It's the "ball" in a game of badmitton that you swat over the net.  It's a weighted rubber half-ball with plastic "feathers" stuck in it allowing it to float more gently to the ground.