an athletic with 1/70 m tall wants to throw a ball to the basket from penalty spot (in basket ball). indicate under which angle and which speed this ball can go to the basket .(x= 5/80)  (h= 3/05)\n(x= the distance between spot penalty till basket)  (h= the distance between ground till basket)
208454
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.
465359
throw a ball against an uneven wall and try to catch it, you wont know where it is going so you need to react fast
351188
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)
583712
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
368213
He would need to throw at least 129 MPH to do that. Here is my calculation:\n\nAssume he throws overhand so that he releases the ball near the ceiling, 8 feet above the floor. Assuming he throws parallel to the floor, the ball starts 8 feet up and begins accelerating down at 32 feet per second square as soon as he releases it. Using the formula X=1/2AT^2, I calculate it takes 0.707 seconds to hit the floor, so it has to travel 133.5 feet horizontally in less than 0.707 seconds to hit the far wall before hitting the floor. 133.5 feet in 0.707 seconds is 189 feet per second or 129 MPH.\n\nBut wait, there is more!\n\nIf he throws underhand, releasing the ball near the floor, and if he throws at a slight up angle so the the ball nearly touches the ceiling half way down the hall, then it takes twice as long to hit the floor. It takes 0.707 seconds to rise up to 8 feet and another 0.707 seconds to fall back to the floor. So he would only have to throw half as fast to hit the far wall; only 64 /12 MPH.\n\nSo the question becomes, can he throw 129 MPH overhand or 64 1/2 MPH underhand?
521290
let ab be the pole,bc  be the shadow \nlet pq be the player then p must lie on ac because \nhis shadow & pole's shadow's meet at the end\nnow trangle abc is similar to pqc(because pq is parallel toab)\ntherefore pq/ab=cq/cb\npq=2m\ncq=1.6m\ncb=1.6+4.4=6m\ntherefore\nab=7.5m\nthanks
109478
Work = Force x Distance\n\nIn this case, since the shot is going up, i.e. opposite the force, the work done on the shot is negative.  Another way to look at it, if up is our positive direction, the distance will be positive but the force, i.e. the weight, is negative.  Note that the distance over which the weight acts is the difference in height, not the entire height of the shot relative to the ground.\n\nW = Force x distance = Weight x distance = - 71.6 N (2.09 m- 1.60m)\n\nI'll leave the arithmetic to you.\n\nTo answer the second part, just notice that the change in potential energy of the shot will be equal in amount to the work done on it.  Just take away the negative sign, since its potential energy is increasing. Another way of looking at it is to say that the work done on the shot takes away its kinetic energy, hence the negative sign, and converts it to potential energy.\n\nSince you're only concerned with the change in potential energy, you only have to concern yourself with the difference in height.  If you want to use the ground as the reference level, and calculate potential energy, you get\n\nDEL PE = PFf - PEi = Weight x final height - weight x initial height\n\nDel PE = 71.6 N (2.09 m- 1.60m)\n\nSo it works out to the same thing.  Yippee.
460430
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.
196316
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
482440
As is usual for these sorts of problems, we'll neglect air resistance.\n\nFor part (b) note that at the top of its trajectory, the velocity of the ball is zero, and the kinetic energy it had when it was first thrown has been convered to gravitational potential energy:\n\n0.5*m*(29.4 m/s)^2 = m*g*h\n\nwhere m is the mass of the ball\ng is the acceleration of gravity (assumed constant over the distance the ball travels)\nh is the maximum height\n\nSimplifying:\n\n432.18 (m/s)^2 = g*h\n432.18 (m/s)^2 / 9.81 m/s^2 = h\n44.07 m = h\n\nFor part a, write the position of the ball as a function of time, (assuming that the ground surface is at y = 0):\n\ny(t) = v0*t + 0.5*a*t^2\ny(t) = 29.4 m/s * t - 0.5 * 9.81 m/s^2 * t^2\ny(t) = t*(29.4 m/s - 4.9 m/s^2 * t)\n\nWe want to know the times when y(t) =0 (i.e., when the ball is at ground level).  Obviously, one of these times is t = 0.  \n\nThe other zero of this equation is found from:\n\ny(t) = 0 = (29.4 m/s - 4.9 m/s^2 * t)\n\n(29.4 m/s)/(4.9 m/s^2) = t\n5.996 sec = t
78428
The basics:\n\nScoring\nA goal between the tall posts= 6points\nA kick into the shorter posts= 1point/behind\nHit one of the tall posts= behind\n\nOther\nA kick must be more than 15m for another player to catch it to be legal or play on\nA tackle should be waste up, neck down\n\nmore info at\nhttp://afl.com.au/default.asp?pg=aflinfosheets&spg=display&articleid=243003
339588
Find the final height of the puck\n\nh = 3 * sin 29 = 1.45\n\nKinetic Energy at bottom must equal potential energy at top.\n\n0.5 m v^2 = m g h\n\n0.5 v^2 = g h\n\nv^2 = 2 g h\n\nv = sqrt (2 g h)\n= 5.33 m/s
554272
If you differntiate the equation you get;\n\ns(t)dt = -32t + 64,\n\nwhen this equation = 0 it gives you the time that it takes the ball to hit it's maximum height. \n\n-32t + 64 = 0  =>  32t = 64  => t = 2.\n\nPlug this into the original equation to get the height;\n\ns(t) = -16*2^2 + 64*2 => -64 + 128 = 64 feet.\n\ni hope that helps.
38492
To further elaborate on what Hakeem said, all players must be at least 10 feet (9.15m) away from the ball during a penalty kick.  The arch ensures this since the area directly behind the penalty spot is less than 10 feet.  All players not involved with the penalty kick must be outside of the penalty area AND the arch.  In addition, all players not involved with the PK must also be behind the ball (no player can be on the side of the penalty area but closer to the end line than the ball).
531933
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.
377577
Yes, use conservation of momentum.  The angle comes into the problem by determining how much of the total momentum is in the x and y directions.  Really, you'll be solving two different equations:\n\nm1*v1x+m2*v2x=m3*v3x\nand\nm1*v1y+m2*v2y=m3*v3y\n\nSo, find each player's initial momentum in the x and y directions, and then solve these equations to find the final x and y velocities of the group.  \n\nGood luck!
390631
You have to take gravity into account as well. \n i can tell you that gravity is approx 32.2 f/s2. So every second, the ball's vertical speed (y component?) will decrease by 32.2 ft/s. \nThis way you can calculate roughly how long it will take before vertical speed reaches 0, after which point the ball will begin to "fall" at the same acceleration rate (32.2 feet per second2). \nWhen this happens you would have calculated exactly half of the ball's trajectory.\nSo if you calculate the distance from the start to when y=0, you can then multiply the distance by two to get the balls total distance, as a ballistic trajectory tends to form a perfect parabola.\nIt has to be noted that my answer ignores wind and air resistance.\n\nForgot to say.\nWhen you ignore wind resistance, the ball's horizontal speed remains constant. Also, find out how high the ball would be 300ft into its trajectory.