How do I determine the max vertical component/height of a golfball after being hit up into the air.?
815859
the vertical component of the velocity applied to the ball is given by\n\nsin 35 = velocity/24.5\n\nvertical velocity is 14.05 m/s\n\nV^2 = u^2 +2gs\n\nat the maximum height, v = 0 so\n\n0 = (14.05)^2 + 2(-9.8)s\n0 = 197.48 + -19.6s\n19.6s = 197.48\ns = 10.07 metres\n\n\nmaximum height the ball reaches is 10.07 metres...
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.
893929
Fg = mg = (0.20 kg)(9.80662 m/s^2) = 1.961324 N\nFc = (mv^2)/r = mrω^2\nFc = (0.20 kg)(0.60 m)((2π rad/s)^2\nFc = 4.73741 N\nFt = √( 4.73741^2 + 1.961324) @ arctan(1.961324/4.73741)\nFt ≈ 5.127 N @ 22.5°\n\n[The vertical component of Ft must offset the weight of the ball.]
427848
I will look for the distance first , if it is 150 yards from the hole I will use 7 or 8 iron . If the wind is blowing towards me I will use 7 iron , If the wind is blowing towards the green I will use 8 iron . Also, if you are facing a short Par 4 (about 280 yards) you can use driver to hit the ball straight to the hole, but becareful, if your driving technique is not good( always slice) then, I will suggest you to use 3 wood then use a sand wedge to chip the ball to the green and putt for birdie . By the way , if you want to test wheather the wind is blowing towards the hole or towards you , you just need to pick up some grass from the fairway or rough , and see how the grass moves in direction, if the grass goes towards you, it means that the wind is blowing towards you, if the grass goes towards the hole, it means that the wind is blowing to the hole, so try this method and give me 10 marks for the best answer .
958247
v=v0-at=0=96-32t\n32t=96\nt=3sec\n\na=32'/s^2\nv0=96'/s\nd=5+96t-16t^2\nd=5+96*3-16*9=5+288-144=5+144=149'\n\nA) The arrow will reach a height of 149'\nit takes 3sec to go up & 3 more to come down to 5'\nthe last 5'\n5=96t+16t^2\n16t^2+96t-5=0\nt=(-96+/-√96^2+4*5*16)/32\nt=-3+/-√9216+320=-3+√9536/32\nt=-3+97.65/32=.05\nB) total time=3+3+.05=6.05sec
446591
1)X=V1t+1/2gt^2\nat the interval between getting to the highest point and being dropped back to the first place is half of the full time = 2 seconds\nV1 at the highest point = 0\nX=displacement\nX=0+1/2*9.8*4=19.6 m \n2)V2=V1+gt\nV2 = velocity of landing\nV1 = velocity at highest point\nV2 = 0 + 9.8 * 2 = 19.6 m/sec
894864
http://en.wikipedia.org/wiki/Darts#Height_and_distance
449240
Atmospheric pressure will support a column of air about 10 metres high.   The exact height varies from day to day depending on atmospheric pressure; high and low pressure systems.   To start with, partly fill your container with coloured water, with the hole at the bottom submerged in more coloured water.   If there was no air in your transparent ball, it would need to be 10 metres in diameterfor you to see a boundary between the empty space at the top and the water column supported by the atmosphere pushing down on the liquid outside, but because you've trapped some air inside, this air pushes down on the water so the water isn't as high.
510674
Its all aerodynamics. Different angle of incidence of air flow on different curvatures give much more flight and movement to a golf ball. This adds more thrill to the play
1033468
2.36 m s^-1 to 2 d p\n\nWe split the launch velocity of the ball into a vertical and a horizontal component.\nHorizontal component = 19.8 * cos(49.1) = 12.9639 m/s\nVertical component = 19.8 * sin(49.1) = 14.9659 m/s\n\nGravity accelerates the ball vertically with the acceleration -9.8 m s^-2 (negative because the pull of gravity is downwards while the ball is moving upwards to start with)\n\nThe vertical distance from the ball to the ground after time t is given by the formula s = u * t + (g * t^2) / 2 where: u = initial velocity, g = acceleration of gravity, t = time elapsed\n\nWhen the ball returns to the ground, s = 0\nTherefore the time taken for the ball to return to the ground is given by\nu * t + (g * t^2)/2 = 0 Substitute values\n14.9659 * t - (9.8 / 2) * t^2 = 0 Remove brackets, divide by t, rearrange\n4.9 * t = 14.9659\nt = 3.05427 s\n\nDuring that time the ball will travel horizontally \ns = u * t where s is distance in m, u is velocity in m s^-1, t is time in sec\nDistance travelled horizontally is 12.9639 * 3.05427 = 39.5953 m\n\nThe player has to run 46.8 - 39.5953 m = 7.20475 m in 3.05427 s\nHis speed must therefore be 2.35891 m s^-1
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
924813
(288^2)/(2*32.2) is the maximum height.\nat time 288/32.2 seconds\nit hits ground at time time   t+576/32.2, where t is obtained from\n45=288+.5*32.2*t^2
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.
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.
326619
heres my two cents, if your hitting it out on the toe of the club, your more than likely topping the ball the other half of the time, 99 out of 100 your clubs are too short, inshort not allowing you to get back to the ball at impact, not that you cant play with shorter clubs, just makes it alittle harder to do it consistantly, go to your local repair shop, have them measure you for the length of golf clubs you should be playing with, i believe the standard length 5 iron is still 37 1/2", goes 1/2" increments from wedges to 1 iron, ie., 6 iron is 37, 7 iron is 36 1/2, you get the point, what they should do is measure you to get the length of club that is best for you, ask if they have a club you can go hit thats your length, go hit a small bucket, that should do the trick, if that helps your ball flight, you have a couple of options, 1 get the guy to put extensions in your clubs and cut to your length, should be about 5 or 6 bucks plus cost of grip, have him measure you for your proper grip size while your at it, just try doing your favorite clubs you use most often first, if that helps do the rest of your set, 2 reshaft your clubs, get ready to pull out your wallet here, can get pricey, then take the advise of the guy above and go thru the full blown fitting process and consider buying new clubs custom built to your golf game. length, shaft flex, lie, grip size, set make-up, if you go and do all this and your still hitting it out on the toe, take advantage of the free deal of the month thing on lessons, and thank god your not at the dam office, grip it and rip it.
299016
Believe it or not, you loft the ball by hitting down on it.  The initial trajectory is determined by the loft of the club face.  Hitting down on the ball gives it backspin, which will actually make it climb higher in flight.  No part of your divot should be behind the ball.  Ideally, the club should strike the ball and ground simultaneously and the entire divot be in front of where the ball was sitting.\n\nUnless conditions were very severe, I would never try to hit up at a ball, even with a driver and the ball on a tee.  That would impart overspin to the ball, which will make it dive into the ground.  Little ball flight and a lot of roll may be called for against a strong wind on a very dry and bald fairway, but is not the preferred shot shape most of the time.  Swing level with your driver.  Let the clubface put the ball in the air.
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)