you throw a baseball straight up at a velocity 64 feet per sec useing s=-16t^2+vt+s how high will it be 1 sec?
400091
1 sec - 48 feet\ntime of max height - 2 sec\nmax height - 64 feet
888567
a) zero momentum, as velocity on top is zero. p=mv\n     p=0.100kg * Om/s\n\nb) Vf = Vi + at\n      0=15+(-10)t\n     t=1.5ec (to reach maximum height)\n\nDivide by 2 for halfway up .i.e 0.75ec\n\nVf = Vi + at\nO=15 + (-10)(0.75)\nVf = 7.5m/s\n\np=mv   (p = momemtum)\n\np = (0.100kg)(7.5m/s)\np = 0.75kg*m/s, upwards (Direction is important)
382082
use d=V1*t + [(a*t*t)/2]\nd=distance=490m\nV1=initial velocity = 0m/s\nt=time = ?\na=acceleration due to gravity = 9.81m/s*s\n\nusing equation above, t will come up to be:\nt=(square root)(2*d/a)\nt=(square root)(2*490m / 9.81m/s*s)\nt=9.99 sec or 10 sec.
894241
answer is approxomately 42.476m/s\n\nonly one formula here \nS=ut + (at^2)/2 \ns- distance travelled = 40m\nu- initial valocity\na - acc due to gravity-9.8m/s^2\nt- time \nin the first case the initial velocity is 0m/s^2\nso\n40=(9.8*t^2)/2\nt=2.857 sec\nas both the balls have to touch the ground at same time  and the second ball being dropped( rather thrown would sound correct) after 2 secs, the later has only 0.857 sec left for traelling a distance of 40m. \nso \n40=u*0.857+(9.8*(0.857^2))/2\nsolving u=42.476m/s\nthe initial velocity for the second ball should be 42.476m/s and should be thrown vertically downwards.\nthat is pheta=0 in the formula given un the question.
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.
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
821062
I think you are supposed to assume the minimum velocity possible, which means tossed at a 45 degree angle. Otherwise he could throw it farther by using the same velocity at 45 degrees.\n\nvelocity = v, y velocity at time t = v/sqrt(2) - 10t so y velocity is 0 at \n\nt = v/(10sqrt(2))\nIn time 2t the distance travelled is 96.90, so\n\n96.90 = 2tv/sqrt(2)\n\nThese are 2 equations with 2 unknowns, solve for t and v (there may be a careless mistake in the details since I did this fast, but the technique will work)
799812
The maximum height is given by the point where the slope (derivative) is zero.\n\nh'(t) = -32t + 96 = 0\n32t= 96\nt = 3\n\nSo the maximum point will be at time t = 3.\n\nh(3) = 5 + 96(3) - 16(3)²\nh(3) = 5 + 288 - 16*9\nh(3) = 293 - 144\nh(3) = 149 ft.\n\nNow just don't stand directly underneath this when the arrow returns to Earth!
353428
0.0161 sec - 0.0115 sec = 0.0046 sec\n\n2 meters / 0.0046 sec = 434.8 m/sec
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.
586983
Impulse = Force*Time = Change in momentum = pf-pi\np=mv\n\nF*t = m*vf - m*vi = m(vf-vi)\n\nF = m(vf-vi)/t\n\nTo find the force behind the hit, you must know how long the ball touched the bat. Also you have to take into consideration that velocity is a vector, so, the final velocity is opposite in sign to the initial velocity.\n\nA baseball's mass is 5oz or 0.141747616 kg.\nSay the velocity that the pitcher throws is 95mph= 42.4688 m/s.\nSay the ball comes off of Barry's bat at -80mph = -35.9m/s\nSay the time of the collision is 0.2s\n\nF = m(vf-vi)/t = \nF = 0.1417kg*(42.4688-(-35.9m/s))/0.2s\nF = 55.5N
831310
It seems the 16 m/s is not relative to this question. That is velocity in the horizontal direction. Gravity only effects in the verticle in this case. \n\ntime = sqrt(2 x height/ g) = sqrt(2 x 45 x 9.8) = 3.03 seconds\n\nvelocity = time x acceleration of gravity = 3.03 x 9.8 = 29.694m/s
723707
s'(t) =(t-3)(-2t^{-3}) + t^{-2}\n=( -2t+6 +t)/ t^{-3}\n=(6-t) / t^{-3}\nso the velocity is 0 when the top =0\ni.e 6-t-0\nt=6\ns
256497
around 1/6th that of earth so around 1.64m/s
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
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.
882625
during the throwing motion the ball is accelerated in the opposite dirrection as gravity. once the ball leaves the hand of the thrower however, the acceleartion is constant, and is equal to that of gravity, g = 9.80 m/s^2