A dart player throws a dart horizontally at a speed of 12.4 m/s. The dart hits the board 0.32m below the?
825231
y=at^2/2\n.32=9.81*t^2/2\n2*.32/9.81=t^2\nt^2=.0652\nt=.2554\n\nx=vt\n12.4*.2554=3.167m =>3.2m
217654
The dart board is hung so that the center of the bullseye is 5'8" (1.74m) from the floor. The throwing line or oche is generally located 7'9.25" (2.37m) from the face of the dartboard measured horizontally.
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
958697
10.4 m/sec\n\nV(t) = (30+a*t) m/sec\n\nwhere a = - 9.8m/sec^2\n\nV(2) = (30 - 09.8*2) m/sec\n\nV(2) = 30-19.6 = 10.4 m/sec
756774
P=P (momentum before = momentum after)\nmv=mv\n(55kg)(4.6m/s) = (230kg)v\nv= ((55kg)(4.6m/s))230kg\nv=1.1m/s
215431
23\n\n0 can be scored by not hitting the board(it could also be argued that 0 is not a true number). 1-20 are regular numbers. 21 is treble 7, 22 is double 11. 23 is the first prime number and can only be divided by itself and 1. therefore not available on a dartboard. \n\nGood lord - dont I sound brainy!!
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
872212
Use the certical component to find the time in the air and since the horizontal velocity does not change; Dx = vx * time\n\nDx = v^2 / g\n\nheight = v^2 / (2g)\n\nHope this helps without actually solving it for you.
829829
Horizontal momentum is conserved.\n\nInitial horizontal momentum is the cats (0) + the skateboard's (1.2 kg * 5.0 m/sec) = 6 m-kg/sec\n\nfinal horizontal momentum is 2.0 kg * s, where s is the speed of the athletic cat.\n\n2s = 6\n\nSpeed s = 3.0 meters/sec
1150360
The acceleration due to gravity is 32.2 ft. per second per second, which is pretty near 32.  Now we need two fundamental equations of physics:\n\nv = a / t\n\nwhich we can re-write\n\nt = a / v\n\nSince a (acceleration) is 32 ft/sec^2, and v (velocity) is 32 ft/sec, time (t) is 32/32 = 1 second until the ball's velocity equals zero.\n\nSo how much distance in 1 second? The second equation gives us the answer:\n\nd = 1/2 a t^2\n\nwe know a = 32, and since we now know t = 1, we can solve the equation. \n\nt^2 = t * t = 1 * 1 = 1\n\nso\nd = 1/2 * 32 * 1\nd = 16 feet
749375
You mean at the same rate - 32.2 feet per second per second.\n\nGravity pulls on all matter at the same rate. If you drop a bowling ball taped to a baseball, they will fall at the same rate. But it is not because the bowling ball is pulling on the baseball, it is because gravity is pulling on both of them. If the tape breaks gravity will still be pulling on both of them at the same rate. They will hit the ground at the same speed, however, they will hit with different forces. Because Force = mass x acceleration. The acceleration is the same for both of them. But the mass of the bowling ball is larger than the mass of the baseball, so it will hit the ground with more force.
862116
KE = 0.5*m*(v)^2 = 0.5*0.023*(848)^2 = 8269.696 J\n\nThis must = work to accelerate the bullet\n\nW = F*d = 8269.696\n\nF= 8269.696/0.803 = 10298.5 N
998671
Gravity slows down the ball at 9.8m/s/s. After 1s, its velocity is 12-9.8=2.2m/s
230718
Answer:\n8 seconds\n\nProof:\nThe problem is really a question of finding what value(s) of x make y equal to zero in the given equation.\nWhen y (the height or distance) equals 0 then the ball is either leaving the ground or hitting the ground. The values of x (the time) that make the height equal to zero tell us when this is happening.\nThe given equation is an inverted parabola so there may be two points where y = 0.\n\n1) Solution 1: x = 0\ny = -16x^2 + 128x\ny = -16*(0^2) + 128*0\ny = -16*0 + 0\ny = 0 + 0\ny = 0\nBut x=0 refers to the time when the ball is first thrown. We want to find the time when the ball lands.\n\n1) Solution 2: x = 8\ny = -16x^2 + 128x\ny = -16(8^2) + 128*(8)\ny = -16*64 + 1,024\ny = -1024 + 1,024\ny = 0\nWhen x (the time) equals 8 seconds, y (the height) equals 0. So ball hits the ground after 8 seconds.\n\nFor your information, after 4 seconds the ball reaches its maximum height of 256 feet.
999681
I'm assuming you meant that 20m down, the two objects collide. Right?\n\nFor the object that is falling down, you know:\n\na = 9.8 m/s^2\nd = 20m\nVi = 0 m/s^2\nVf = ?\nt = ?\n\nUsing the formula: Vf^2 = Vi^2 + 2ad\n\nYou can find that: \n\nVf = 19.798989....m/s^2\n\nYou plug that into the next formula to find time (t), which is: \nVf = Vi + at\n\nYour answer should be t = 2.020305......s\n\nYou use that number for the second object following the formula: d = vt (because the horizontal velocity will never change since there is NO horizontal acceleration)\n\nd = 2.5(2.0203....)\n\n= 5.05m\n\nI hope there aren't any mistakes! ;D
221114
I understand what you're saying, but you forgot one small thing:\n\nThe bowling ball has more mass, and therefore needs more pull to reach the same speed because of momentum. You know, it's harder to push a car than push a cup. The momentum pull and the pull of gravity completely cancel each other out, so all things fall at 9.81 m/s/s.
409599
The Official throwing distance, for most countries, is 2.37 meters, as measured along the floor, from the plane of the face of the dartboard. In feet, this is 7 feet, 9-1/4 inches. The height of the board, to the center of the bull, is 173 centimeters, or 5 feet, 8 inches.\n\nCheck the link if you need more info!