If the downward velocity of the leg is 6.7 m/s when the joggers foot hits the ground, and if the leg stops . .
91548
tried this problem, but when converting 2cm disance to meters, the acceleration was way out, too large.I used the equation V2 = U2 + 2aS to arrive at the acceleration (deceleration). This might have been the wrong approach.
786359
In theory, it will never stop, so I'll assume that when it bounces less than 1 mm it is finished.  Total distance travelled will be 26.99221155 meters.  Start with the 3 m drop and add the 2.4 m bounce and drop and continue the sequence until the bounce is less than 1 mm.  Here's my spread sheet:\n\n3000----------3000\n2400----------7800\n1920---------11640\n153614712\n1228.817169.6\n983.0419135.68\n786.43220708.544\n629.145621966.8352\n503.3164822973.46816\n402.65318423778.77453\n322.122547224423.01962\n257.698037824938.4157\n206.158430225350.73256\n164.926744225680.58605\n131.941395325944.46884\n105.553116326155.57507\n84.4424930126324.46006\n67.5539944126459.56804\n54.0431955326567.65444\n43.2345564226654.12355\n34.5876451426723.29884\n27.6701161126778.63907\n22.1360928926822.91126\n17.7088743126858.32901\n14.1670994526886.6632\n11.3336795626909.33056\n9.06694364726927.46445\n7.25355491826941.97156\n5.80284393426953.57725\n4.64227514726962.8618\n3.71382011826970.28944\n2.97105609426976.23155\n2.37684487526980.98524\n1.9014759 ------------26984.78819\n1.52118072 -----------26987.83055\n1.216944576----------26990.26444\n0.973555661----------26992.21155\nOk, Excel columns don't work well when copied here, so I've added spaces to a few entries at the top and bottom.\n\nGuess spaces didn't work either.  Maybe the dashes will stay.
105355
the change in speed. you are walking at 6kms/hr but because you get tired, your speed became 4kms/hr. your acceleration is minus 2kms/hr. if you change your speed from 6kms/hr to 9kms/hr, then your acceleration is posituve 3kms/hr
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
164406
the distance separated the runners is dx= v2.dt where v2 is thespeed of runner  2 such that v1= 1069/3:42.46 = (calculate it) ,v2=1069/3:52.72= (caculate it) the difference in time is 3:52.72-3:42.46=0:10.26mincalculte - then calculte the distance dx= v2.dt  subsitute you get the solution
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.
219948
walking requires one foot to be on the ground at all times... running- both feet are off.... so, the instant there is that lapse where both feet aren't on the ground
812369
Sorry, no.  You'd splat at the bottom of the hill in real life.\n\nThe reason for this rests in a physical law that says if you're going straight, you'll continue straight (down hill) until a force stops you from going straight.  Spinning is not that force.\n\nThe momentum (which is the thing that keeps you going straight) is linear (straight).  Spinning is angular (it goes around, not straight).  So the forces required to start spinning are not straight (they act over angles) and will not affect your straight march into that wall at the bottom of the hill.\n\nOf course, if you simply fall down, the ground pushing on you as you hit it is a linear force.  Your linear momentum will be changed (stopped in this case).  Thus, it will in fact stop you on your downward destiny with that wall at the bottom of the hill.\n\nI tried to keep equations and such out of this, just the one notion: momentum will keep you going straight downhill until you apply a linear force (not angular like spinning) to stop your run.
870847
work out your legs and keep running\nb
815859
the vertical component of the velocity applied to the ball is given by\n\nsin 35 = velocity/24.5\n\nvertical velocity is 14.05 m/s\n\nV^2 = u^2 +2gs\n\nat the maximum height, v = 0 so\n\n0 = (14.05)^2 + 2(-9.8)s\n0 = 197.48 + -19.6s\n19.6s = 197.48\ns = 10.07 metres\n\n\nmaximum height the ball reaches is 10.07 metres...
931347
x =70m, t = 7s, Vf = 15m/s\na.)\nx = 1/2(Vi + Vf)t\nVi = initial velocity\nVf = final velocity\n\n70m = 1/2(Vi + 15m/s)(7s)\n10m/s = 1/2Vi + 7.5m/s\n2.5m/s = 1/2Vi\nVi = 5m/s\nb.)\nVf = Vo + at\n15m/s = 5m/s + 7s(a)\n10m/s = 7s(a)\na = 1.43m/s/s\nbasic kinematics\nyou can't use distance = rate x time because there is acceleration and the speed is not constant.
12020
The velocity will be 0 when it reaches its highest and lowest point
860846
Force = 60 N.\nMass = 3 kg\nTime = 10 s\n\nAn object's change in momentum, Δp equals the Impulse, I on the object. I = ∫F dt = Ft. \n\nΔp = I = Ft = (60 N) * 10 s = 600 Ns\n\nAssuming the object started at rest,\np (initial) = mv = (3 kg) (0 m/s) = 0 Ns\nΔp = 600 Ns\np (final) = 600 Ns\n\np (final) = mv = (3 kg) * v(final)\nv(final) = (600 Ns) / (3 kg) = 200 m/s\nΔv = v(final) - v (initial) = 200 m/s - 0 m/s = 200 m/s\n\nThe object's change in momentum was 600 Ns (kg * m/s).\nThe object's change in velocity, assuming it started from rest, was 200 m/s.
806819
A.\n\nF=ma, so, a= F/m\n\nIn this case, a=7/5ms^2\n\nB.\n\nv=u+at, u=0 (initial velovity), t = 2 sec, so\n\nv=0+7/5 x 2 metres per second = 14/5 m/s\n\nC.\n\nThe time to reach Sasha will be:\n\nt = d/v = 11.2/(14/5) seconds\n\nI will let you use your calculator!
232038
Distance = rate x time\n\nif you're going 8 feet per second for 4 seconds, you've gone 32 feet...
919282
please make your question more understandable\n\nif it decreases from 30mph to 15mph over 75 metres then at this deceleration its logical to say it would take another 75metres to stop\nbut something tells me thats not the answer ! if you want me to do your homework at least have the decency to type in the full question.
409505
A soccer ball of mass 0.43 kg leaves the foot of the kicker with an initial speed of 25 m/s.The weight of the ball at the start of the game shall not be more than 450 g nor less than 410 g.\nSource-Sam