you throw a rock straight up at 30m/s, how fast will it be moving 2 sec later?  No friction?
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
970380
Let rock be thrown at an angle of θ at a velocity of Vo\n\nThen vertical component = Vo sinθ\nand horizontal component = Vo cosθ\n\nAlso acceleration in horizontal direction = 0 \nand acceleration in vertical direction = -g = -9.8 m/s²\n\nVertical direction\n\na = -9.8\nv = u + at\nSo Vt = Vosinθ - 9.8t\nand s = ut + ½at²\nSo h = Vosinθ*t - 4.9t²\n= 0 when it is at ground level\nie t(Vosinθ - 4.9t) = 0\nSo t = 0 or Vosinθ/4.9 ... ****\nSo it will take Vosinθ/4.9 s to cross the river.\n\nHorizontally\n\nVt = Vocosθ + 0 (as a = 0)\n= Vocosθ\n\ns = ut + ½at²\n= Vocosθ * t\n\nSo 75 = Vocosθ *Vosinθ/4.9\n= Vo².sinθcosθ / 4.9\n= Vo² * 2sinθcosθ / 2*4.9\n= Vo² * sin2θ/9.8 \nThus Vo² * sin2θ = 75*9.8 = 735\nThus Vo² = 735/sin2θ\nNow least value of Vo occurs when Vo² is minimum and this will occur when sin2θ is maximum (=1)\n\nSo Min possible Vo = √735 \n≈ 27.11 m/s (projected at 2θ = 90° ie θ = 45°)\n\nMinimum time = Vo*sin45°/4.9 (from ****)\n≈ 3.91 s
10186
Depends on the media in which it propagates. The temperature and the pressure also slightly affect the speed. In air it's about 330 m/s.
754389
Learn to answer these questions yourself!\n\nIf the air is assumed to be frictionless, the vertical and horizontal velocities are independent.  The altitude is given by H = H0 + V0t + 1/2 a t^2.  H0 is +30 meters, V0 is +24.5 m/sec, and a = -9.8 m/sec^2.  If the target is at ground level, you are solving a quadratic for H = 0.\n\nSolving the quadratic gives you the time of flight, t.  Multiply the time of flight by the horizontal velocity (49 m/sec * cos(30°)) and  you get the distance of the target.\n\nOf course, you can just look at the problem and the answers and see if there's just one answer that's anywhere close.  In this case, the initial vertical speed is 24.5 m/sec; it takes 2.5 seconds for the arrow to rise to its peak altitude, and another 2.5 seconds to come back down to the altitude of the bow (5 seconds total).  Falling the final 30 meters would take on the order of a second.  The cosine of 30 degrees is about 0.87, so the horizontal velocity is about 42.5 m/sec; multiply by 6 seconds, and you get about 250 meters.  Ergo, (4) is the only answer that's even close.
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
520357
Try it at different speeds,\nEx:-\nFirst try at 2 m/s\nThen try at 5 m/s\nThen try at 10 m/s\nThen if you survive, try at 20 m/s\nTry using concrete wall if you want to have more fun :)
882649
First solve for the force of friction between the block and the table using the formula,\n\nf=uR, where f is the force of friction, u the coefficient of static friction, and R is the force normal to the table. R therefore is the weight of the block equal to 31g.\n\nf=0.425*31*9.8\n =129.1N\n\nThe weight of the sand added to the bucket plus the weight of the bucket itself should be just slightly equal to 129.1N so that the system will begin to move. At that point, however, the block will begin to accelerate because the force of friction acting on it will no longer be static friction but kinetic friction, which is lower in magnitude. Thus there will be a net force equal to 129.1 minus the force of kinetic friction on the block. Thus,\n\nNet force, F=129.1-0.320*31*9.8\n                 =129.1-97.2\n                 =31.9N\n\nF=ma, where F is our net force of 31.9N, m the total mass of block plus mass of sand and bucket, and a is the acceleration of the system.\n\nMass of sand and bucket=129.1/9.8\n                                    =13.2kg\nSubstitute known values:\n\n31.9=(31+13.2)a\na=31.9/44.2\n  =0.72m/s^2
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
794040
This is really a calc question.\n\nAn item will fall at 32ft per sec per sec from a rest.\nh - the free fall = t\n\n2. As it is stated the axle is frictionless, therefore the cyinder and axle will move and there is no force between the two. However there is a force between the rope and or bucket and the cylinder.\n\ndouble check the 32ft/s in your book, there may be more to the equation.\n\nCarl
747360
Well gosh....  \n\nsince the only linear unit mentioned in the question is 18 ft/s  let's assume that the 60 is going to give us acceleration in ft/sec^2 \n\nSo, the position function for this is the second integral of the acceleration function, right? \n\nso...  the first anti-derivative of 60t is  30t^2   which gives the velocity, so at the end of the boost phase, our rocket is moving 30 * 3 * 3 = 270 ft/s \n\nand the second antiderivative is 10t^3  right?  \n\nSo, the position at time T is 10 * 3 * 3 * 3 = 270 feet.  (odd coincidence, but there you are...) \n\nNow, the rocket begins a SECOND acceleration by gravity of -32 ft/s negative 'cause we're accelerating DOWN now... ....   but this time the acceleration is constant,  so \n\nV = gt   and s = 1/2  a  t^2 so, we're slowing from 270 feet per second to zero at the top...  and  t = v/g = 270 / 32 = 8.44 seconds and the height we coast to is the 270 we had at the end of boost plus   1/2  *  32  * 8.44 * 8.44 = 1140 feet.  \n\nThe parachute is going to open after 14 seconds...   so, 14-8.44 is 5.56 seconds of free-fall (and we'll ignore air resistance again) so, we go from zero to something in 5.56 seconds....   and position is 1140 - ( 1/2  * g  * t^2 ) again...   so, the parachute opens at \n\n1140 - (1/2  * 32  * 5.56 * 5.56 )  = 645 feet. \n\nad the velocity is  gt = -32 * 5.56 = -178 feet per second...  \n\nNow, for the NEXT five seconds, we slow under the effect of the parachute   to  -18 ft/sec -- so,  we slow by 160 feet per second in five seconds, so the acceleration of the parachute is (oddly) 32 feet/second squared (huh, how odd is that? )  \n\nSo, the distance traveled during that phase is Vo t + 1/2 at^2  \n\nSo, we have (-160 * 5) + (1/2*32*5*5) =  -800 + 400 = 400 feet.  \n\nand from 400 feet up, we continue down at 18 ft/s for an additional  22 seconds...  \n\nand the rocket lands at t = 22 + 5 +  14 + 3  = 44 seconds \n\nAnd yes, I dropped a bunch of decimal points.  Your original data was only to two significant digits....
438060
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
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
698763
Vy=sin(36)(30)=17.63m/s\nVx=cos(36)30= 24.27m/s\n\nwhen it reaches the ground, the speed will be the same as Vy\nVf=at+Vi\n-17.63m/s=(-9.8m/s^2)t+17.63\n-35.26m/s=(-9.8m/s^2)t\nt= 3.597 s\n\nwhen it reaches the highest point, its velocity will be 0m/s\nVf^2=2ad+Vi^2\n(0m/s)^2=2(-9.8m/s^2)d+(17.63m/s)2\n-310.81m^2/s^2=(-19.6m/s^2)d\nd= 15.85m\n\nx=vt\nx=(24.27m/s)(3.597s)\nx=87.29m
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
932992
it is a much easier problem than it seems..\nthe key point is the fact that the body is moving at constant speed. meaning that the net force on it is zero --> no acceleration\n\nthere are two forces. friction and the horizontal component of the force that is exerted at 40 degrees.\n\nso simply ...\n\n\nFx = Friction==>\n976 cos40 = coeff. *N *g\n\n\n now there is equilibrium at the vertical axis as well.\n\nmeaning that \n\nN+Fy = m*g =>\n\nN = (mg- fy)\n\nso we mix them together and hve\n\n976cos40 = 0.667 (mg- 976sin40) =>\nm=50.31kg\n\nin this problem you use that there is equilibrium in both of the x and y axis.
979735
If we just give you the answer, you'll never learn properly.\n\nAs has been suggested you need to get everything in the right units first!!\n\nNewtons 2nd law F= Ma\n\nForce = Mass X Acceleration\nForce is in Newtons\nMass in Kilograms\nAcceleration in Metres per second\n\nYou already have the mass - 38,000kg\n\n\nYou need to figure out the acceleration:\n\n\n60 kilometres/hr = 60,000metres/hr\n60,000metres/hr divided by 60 = 1,000 meters per minute\n1,000 metres per minute divided by 60 will give you the needed finishing velocity of metres per second\n\nAcceleration = finishing velocity / time ( 20 seconds )\n\nForce therefore = 38,000 X your answer for acceleration!
873979
deceleration = change in speed / time\ndeceleration = (50 (km / hr)) / (0.15 s) = 92.5925926 m / s2\n\nF = ma\nF = (1 500 kg) * 92.5925926 (m / (s^2)) = 138 888.889 newtons