a ball is thrown vertically...?
438060
Without regard to air resistance you solve the problem as such.\n\nv(f) = final velocity\nv(i) = initial velocity\na = acceleration\nt = time\nx = displacement (distance)\n\nWhen you throw the ball up vertically, you know several things:\n1) v(i) = 30 m/s\n2) v(f) = 0 m/s\n3) a = -9.8 m/s^2 (this is because acceleration due to gravity goes in the opposite direction of the object)\n\nSolve for x with this equation:\nv(f)^2 = v(i)^2 + 2ax\n0^2 = (30 m/s)^2 + 2(-9.8 m/s^2)x\n0 = 900 m^2/s^2 - (19.6 m/s^2)x\n(19.6 m/s^2)x = 900 m^2/s^2\nx = (900 m^2/s^2)/(19.6 m/s^2)\nx = 45.9 m (rounded off)\n\nNow you know how high the ball went.  The time it takes to reach this is equal to the time it takes to get back down (assuming that it is the same distance up and down).\n\nTo solve for t, use this equation:\n\nv(f) = v(i) + at\n\n0 = 30 + (-9.8)t\n9.8t = 30\nt = 30/9.8\nt = 3.06 s  this is going up to max height.\n\nto check if it will be the same time going down, use this equation:\n\nx = v(i)t + (1/2)at^2\nv(i) = 0 (because when you reached max height, your velocity will be zero)\na = 9.8 because your ball is travelling in the direction of the acceleration.\n45.9 = 0 + (1/2)(9.8)t^2\nt^2 = 45.9/4.9 = 9.367\nt = 3.06 s\n\ndone
523948
This happens when the baseball player has the option of getting the out in multiple places.  For example, a ball is hit to the short stop and there was a man on first.  In this case, the short stop could throw the ball to second or first base and get the out.
1075352
a ball n some players lol
803160
If they are thrown with the same MAGNITUDE of velocity, and you are either not counting air resistance or are assuming that they all have the same drag parameters, then the MAGNITUDES of the final velocities will all be the same.  The actual velocities of the ones thrown up and down will be exactly the same while the one thrown horizontally will have a horizontal component an a lesser vertical component.  Still, all magnitudes will be the same.\n\nIf however, they are thrown at different speeds, then it will be the one that is thrown the hardest If you are counting air resistance and the balls have different drag parameters, then you have not provided enough info to tell.\n\nAnd if by final velocity you mean after they have come to rest on the ground, then obviously they all have a velocity of 0.  Typically this is not referred to as final though.
630174
It hasn't always been hexagons.  The balls used to be made from curved four sided panels.  Using hexagons ensures that the ball is more spherical.
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.
1053965
Simple parabolic path formula \n\na hit ball is given a velocity v at and angle theta.\n\nAssuming no air resistance the balla travels through the air along its horizontal vector un abstructed, while acted on by gravity in the vertical direction.\n\nthe equation\n\nd=vt + 1/2gt^2 can be solved for d which will be the total vertical distance at time t, where t is the time at which the ball is the highest, lets say about 1/2 of the total time, g is the gravitational constant and v in the velocity in the vertical direction
934381
Basically you know the initial velocity that the baseball starts out with in both x and y.  Let V = 25m/s.  For A:  Vxo (initial velocity in the x direction) = 25*cos(60).  Vyo (initial velocity in the y direction) = 25*sin(60).  You know this similarly for ball B with a different angle.\nYou can use your kinematic equations\nd = Vo*t + (1/2)*a*t^2\nVf^2 = Vo^2 + 2*a*d\nVf = Vo + a*t\nd = (1/2)*(Vo+Vf)*t\nin both the x and y directions to determine whatever quantities you want.  \nUse the knowledge that at height yA, the velocity in the y direction is 0.  Similarly for ball B.  \n\nThat should get you started.  Good luck!
434509
Here is what you want to know. Toss an object into the air. The point at which it changes direction is the point at which the gravitational force on it is exactly equal to the force you applied when you threw it. This is the maximum height. At that point only is it "weightless" immediately before and after it is under the effects of gravity.
369849
i think the ball nd basket height is a lil shorter....not sure....
74061
the simple answer is if you were to throw a ball at the groud 15 feet away from you the angle created between the ground and the path of the ball when its traveling towards the ground is teh angle of incidence\n\nhope this helps
888456
Mathematically, on the moon you could now throw the ball 240m.\n\nScientifically, on the moon you could throw the ball 240m and farther. On the moon, their is only gravity bringing the ball to a stop. On Earth, gravity and friction of the air slows the ball down
1086779
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.
998671
Gravity slows down the ball at 9.8m/s/s. After 1s, its velocity is 12-9.8=2.2m/s
698592
The acceleration due to gravity is independent of the mass of the object.\n\n The speed of any mass thrown up will decreases by an amount 9.8 m/s per second. \n\nIf the initial speed is u, then its kinetic energy is ½ m u u. Its potential energy is zero.\n\nThe total energy is½ m u u + zero = ½ m u u.\n\nAs it goes up its kinetic energy decreases whereas its potential energy increases keeping the total energy same.\n\nWhen it goes up through ‘h’ meter,  its total energy is [ ½ m u u + mgh].\n\n[ ½ m u u + mgh] = ½ m v v.\n\nm u u + 2mgh = m v v. \n\nCancelling m on both sides,\n\nu u + 2gh = v v\n\nThus we have the familiar equation vv = uu + 2gh.\n\nWhen the ball reaches 9 m its speed becomes zero.\n\nHence,  v v = 2 x 9.8 x 9\n\nSolving we get v = 13. 3 m/s . \n\nThe speed with which it is thrown up is 13. 3 m/s.\n\nWhen the height is 8 m, now having known the initial speed we use the same equation for solving for uu.\nvv = uu + 2gh\n\n13.3 x 13.3 = uu + 2 x 9.8 x 8 .\n\nSolving for u, we get  u u = 196 or u = 4.43 m/s.\n\nThe problem can be viewed in other way.\n\nWhen the ball is at its heighest point namely at 9m its potential energy is mgh. where h is 9m\n\nWhen it falls through one meter its kinetic enery increases and potetial energy decreases , keeping the total energy as the same.\n\nmg x 9 = ½  m vv + mg x 8.\n\nOr mg ( 9-8) = ½  mvv \n\ng x 1 = v  v/2\n\n2 g = v v. \n\nv v = 2 x 9.8 = 19.6 (m/s)^2. \n\nv = 4.43 m/s.\n\n Note that the mass gets cancelled and hence we need not know the mass of the ball.
575287
It's not really a trick question, if the answer is just "throw it straight up," but I can tell you from experience that throwing a ball straight up as hard as you can does NOT get as much velocity as throwing it semi-horizontally; it's just too awkward of a motion, you can't quite get the same force behind it, since you can't use your body as effectively.\n\nA way that you CAN throw a ball as fast as you're capable of, and still catch it, is to throw it against a hard surface that it will bounce off of, so that it comes back to you.  Works great with tennis balls, not so well with baseballs, though they do bounce some...
664646
just make it really tall, cause the ball goes pretty high but then again it cant do any real damage to it.