A 46.9-g golf ball is driven from the tee with an initial speed of 52.4 m/s and rises to a height of 28.5 m?
816137
a) v^2-u^2=2as\n    v^2-52.4^2=2*(-9.8)*28.5\nv^2= 2745.8-558.6\n     =2187.2\nv=46.76m/s\n\nKinetic Energy=1/2mv^2\n                     =1/2*46.9*46.76^2\n                      =51,290 Joules\n\nb)height=28.5-5.06=23.44m\n\nv^2-u^2=2as\n\nv^2-52.4^2=2*(-9.8)*23.44\nv^2=2745.8-459.42\n     =2286.4\nv=47.8m/s
127076
(a) 86N / 45kg = 1.91m/s^2\n(b) 15kg * 1.91m/s^2 = 28.67N\n(c) 15kg * 1.91m/s^2 = 28.67N\n(d) 15kg * 1.91m/s^2 = 28.67N\n(e) 86N * (2/3) = 57.33N\n(f) 86N * (1/3) = 28.67N\n(g) Yes
573834
Its potential energy was 78.4 m joules before falling, where m is the mass of the ball.\n\nAfter collision when it reaches the height 5m its potential energy is 49 m joule only.\n\nIt has lost 8m joule of energy due to inelastic collision.
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.
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.
759333
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m
905071
The trick to solving this problem is to set up your equations properly:\n\nR = Vx * t\n\nH = Vy * t - (1/2)g * t²\n\nWhere R is the projectile's range and H is its initial height, Vy is the projectile's initial vertical velocity and Vx is its horizontal velocity.\n\n[[sorry for the edit, I just noticed the 45° angle; that changes the analysis]]\n\nHere's what we'll do: let's flip the problem around and pretend that instead of landing at a downward angle of 45°, it takes off at an upward angle of 45° with some velocity V, and at some point, t, the projectile has the coordinates (50, 40).\n\nThis introduces two new equations:\n\nVx = V * cos(theta)\nVy = V * sin(theta)\n\nCombining the equations now provides:\n\nR = V * cos(theta) * t\nH = V * sin(theta) * t - (1/2)g * t²\n\nSolving for t in equation 1:\n\nt = R / (V * cos(theta))\n\nSubstituting this into equation 2:\n\nH = V * sin(theta) * R / (V * cos(theta)) - (1/2)g * (R / (V * cos(theta)))²\n\nH = R * tan(theta) - (1/2)g * (R / V)² * (1/ cos(theta))²\n\nWe can now solve for V as:\n\n-H * cos(theta)² + R * tan(theta) =  (1/2)g * (R / V)²\n\n[-H * cos(theta)² + R * tan(theta) * cos(theta)²] =  (1/2)g * (R / V)²\n\nV² = (1/2)g * R² / [-H * cos(theta)² + R * tan(theta) * cos(theta)²]\n\nV = [(1/2)g * R² / [-H * cos(theta)² + R * tan(theta) * cos(theta)²]]^(1/2)\n\nSubstituting the values provides:\n\nV = 49.497 m/s\n\nFinally, plugging this back into the expression for Vx:\n\nVx = V * cos (theta)\n\nVx = 49.497 * cos (45°)\n\nVx = 35 m/s\n\nIf we now, flip our coordinate system back around, we know that the horizontal component of velocity is 35 m/s (which does not change over time). However, this implies that there will be some vertical component of velocity when the projectile is fired. We find this by knowing that the final vertical component Vy equals the final horizontal component (to form the 45° angle):\n\nVx = Vy\n\nAdditionally, the vertical component of velocity changes due to gravity. Thus, the initial vertical velocity component is:\n\nVyo = Vy - g*t = Vx - g*t\n\nWhere t = R/Vx = 50/35 = 1.42 s. Plugging this in, we get:\n\nVyo = 21 m/s (downward)\n\nFINAL RESULTS:\n\nVx = 35m/s\n(Horizontal Velocity)\n\nVy(t) = -21 - 9.8*t\n(Vertical Velocity as a function of time)\n\nH(t) = 40 - 21*t-(1/2) * g *t²  \n(Projectile Height as a function of Time)\n\nR(t) = 35 * t\n(Range as a function of time)\nHope that clears things up a bit! :)\n\nVo = sqrt(Vx² + Vyo²) = 40.82 m/s\n(Launch Velocity)\n\nLaunch Angle = arctan(Vyo/Vx) = -30.96°\n\nTime of Flight = 50/35 = 1.42 seconds\n\n-----------\n\nYes, it's long, I know. But I hope it clears stuff up for you :)
702948
asume that it is free fall that is initial speed is zero\n\nv = u+at = 0 + 9.8 * 5 = 49 m /sec\n\nit is independent of weight of stone
909202
a= r * w^2 = .12 * 2^2 = .12 *4 = .48 m/s^2
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.]
1120113
The miximum height reached by the ball is when the velocity of the ball is zero.  Since the initial velocity is 64ft/s, the distance above the ground equation becomes s = -16t^2 + 64t + 0\ns = -16t^2 + 64t .  Taking derivaties to find velocity equation.\nds/dt = -32t +64.  At maximum height ds/dt = 0.  \nThus, -32t + 64 = 0\n            32t = 64\n              t = 2\nSo, maximum height is reached at t = 2s.  So the maximum height reached by the ball:\ns = -16(2)^2 + 64*2\ns = -64 +128\ns = 64 feet.\nThe maximum height reached by the ball is 64 feet.
695335
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
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!
840248
Vertical components\n\nv = u + at\n0 = u + (-32)(2.1s)\nu = 67.2ft/sec\n\nMaximum height the ball reached\n\nv^2 = u^2 +2gs\n0 = (67.2)^2 + 2(-32)(s)\n64s = 4515.84\ns = 70.56\n\ndownward velocity 0.8 seconds after the ball reached the max.height.\n\nv = u +at\nv = o +(32)(0.8)\nv = 25.6 ft/sec\n\ndistance from the maximum height\n\nv^2 = u^2 +2gs\n(25.6)^2 = 0 + 2(32)s\ns = 10.24 ft\n\nHeight of the fence = 70.56 ft - 10.24 ft\n\n\nHeight of the fence is 60.32 ft...\n\nthere ya go...
1140714
Bigger than a golf ball, smaller than a baseball.
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.
643339
It's all about conservation of energy.   The contestant did some work on the ball.   Remember the definition of work?\n\nWork = Force * Distance\n\nSo, the contestant did a work:  \n              W = 810N * 2.0m = 1620Joule\n\nOk, since energy is conserved, the ball will have a kinetic energy of 1620Joule at the begining of the throw, but it will eventually lose it as it goes up, transformed into Potential energy.\n\nWhen the ball reaches the highest point its speed is equal to zero, and hence the kinetic energy is zero too.  Conservation of energy requires that all 1620Joules transformed into potential energy.  So, do you remember the formula for potential energy?\n\nP. energy =  mass * gravity * height ; \n\nwith gravity = 9.82m/sec^2\n\nSince the potential energy is equal to 1620Joules:\n\n1620Joules = (7.0Kg)*(9.82m/sec^2)*height\n\nSo:\n\nheight = (1620Joules)/(7.0Kg * 9.8m/sec^2) = 23.567m\n\nHope I helped