If the downward veloc of the leg is 6.4 m/s when the joggers foot hits the ground, and if the leg stops in?
111044
I'm just doing all your work aren't I?  Its ok I like physics, and its good practice.  I am studying for the MCAT coming up this April (Med School).  This problem unlike the other ones that I did is a little more tricky to me.. Mostly because I don't have my physics textbooks with me.  But it would appear that you need to calculate the acceleration that is exhibited on the foot when it hits the ground.  The way I approached this problem is by using vf^2 = vi^2 + 2ad  and I solved for acceleration.  Knowing that the final velocity is zero I isolated acceleration to get and acceleration of -1024 m/s^2!!! thats huge isn't it.  that is why I'm uncertain about this problem... anyways, multiplying that using the F =ma equation I got a total force of 9216N.  But after inspecting the inital velocity its got to be right because if you convert 6.4 m/s into km/hr you get 23.04 km/hr!!! This person must be falling of a building or something!  So there it is.  Afterall didn't the question start out saying that injury can happen!?
317623
i believe the formula is\n\n16 foot per second per second\nor\n16ft + 32ft + 48ft + 64ft + ect....\n1 sec, 2 sec, 3 sec, 4 sec, ect....\nadding 16 foot for each additional second\nuntil you are adding the terminal velocity for each second.\nthe terminal velocity is the fastest something can fall.\nthe terminal velocity changes with the shape and mass\nof the falling object.\nexample : a piece of paper vs a sphere of lead\nyour ball would be approx. 96 ft of the way down\nor 4 ft off the ground, provided it wasn't Styrofoam.
758596
Initial velocity = Vo = 32 ft/sec\nInitial height = So = 0 ft (starting from the ground)\n\nV at time t = Vt = Vo - 32t \n(each second the baseball goes 32 ft/sec slower)\n\nS at time t = Vot -16t^2\nS at (1 second) = 32 ft - 16 ft = 16 ft.\n\nS(maximum) occurs when the Velocity = zero (before it starts downward)\nV = zero when Vo - 32t = 0\n32 = 32t, t = 1 second\n\nThe maximum height is 16 feet, which happens at t = 1 second.
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)
295799
k ill try to help u  even though i dont condone cheating. i just know how it is to not understand something and its due in a few hours and it sucks. just remember, nobody will be able to help u on the exam.\n\nso 1. \na. acceleration is change in velocity divided by time. \nsince it started from rest, initial velocity is 0 and then final velocity is given as 25.3 m/s. and the time is given as 0.59s. so to get acceleration   you would do  \na =  Vf-Vi / t\n  =  25.3 - 0 / 0.59\n  = 42.88......\n\nb. the force is equal to mass x acceleration. since you know the mass and acceleration, you just multiply them and you get F = 38421.69... N. \n\nc. you have to find the normal force in this case, which is the force that opposes gravity. like upward force. so in this case, gravity is acting on the driver and force of gravity is equal to mass x 9.8 = 666.4 N. and that means the normal force is 666.4N (since he is not moving up or down the total vertical force = 0)\n\n2. ok well in this case, you know the initial velocity is 0, the distance is 404m and the time is 4.944s. if its constant acceleration, you could assume that the average velocity is just Vi+Vf/2. so if thatz average velocity, then distance = average velocity x time\n     404 = 4.944s  x  (0+Vf)/2\n 404 x 2 / 4.944 = Vf\nVf = 163.43........m/s\n\nthe acceleration would just be Vf-Vi/t. and since you know Vf now  you just do   a = 163.43-0/4.944\na = 33.056....m/s^2\n\n3. ok well acceleration is 9.8m/s^2 (gravity)     distance is 9.7m. initial velocity is 0m/s. so this time, you have to solve 2 equations. the first one is the acceleration equation. a = Vf-Vi / t = Vf / t = 9.8\n9.8 = Vf/t\nVf = 9.8t\nand the other one is the distance = t  x  Vi+Vf/2   average velocity one. \n9.7 = t x Vf / 2\n19.4 = Vf x t\n19.4 = (9.8t) x t\n19.4 = 9.8t^2\nt^2 = 1.979...\nt = 1.406....s\n\nnow that you know the time, you find the Vf by putting it in any of those 2 equations. Vf = 9.8t = 9.8 (1.406)\nVf = 13.788... m/s\n\npart b. so in this case, you use the initial velocity as the one u just found (the velocity of the swimmer at the surface of the water)   so Vi = 13.788m/s\nand Vf = 0    since they come to a stop. and d = 2m\nso you have to find the acceleration. first you need to find the time by using the average velocity equation.\nd = t x Vi+Vf/2\n2m = t x (0+13.788)/2\n4 = 13.788t\nt = 0.29s\nnow that you know the time, you have to find acceleration\na = Vf-Vi/t\n  = 0-13.788 / 0.29\na = -47.5286m/s^2    its negative because its downwards and up is positive in our case. so the force = mass x acceleration and you just got acceleration and her mass is 65kg so you just multiply by 65kg to get F = 3089.36N\n\n\n4. well F = ma and you have the force. when they say it weighs this much newtons, then it means  use gravitational acceleration which is 9.8. so 2350N = m x 9.8m/s^2\nm = 2350/9.8\nm = 239.79kg\n\ni hope you actually understand this stuff. try to do your homework by yourself   dont take advantage of this site. youll just end up screwing yourself over. and if u dont get anythin, let me know and ill try to help
634973
Basic physics... let me look up the formula:\n\nRange = (v cos A / g ) ( v sin A + sqrt(v^2 sin^2 A) )\n\nBut 10 mph = 52800 feet / 3600 seconds = 14 2/3 feet / second.\nSo:\nv = 14.667 feet/sec\ng = 32 feet/sec^2\ncos (60) = 1/2\nsin (60) = sqrt(3)/2\nsin^2(60) = 3/4\n\nRange = (v cos A / g ) ( v sin A + sqrt(v^2 sin^2 A) )\nRange = (14.6667 * 0.5 / 32) (14.6667 * sqrt(3)/2 + sqrt(14.6667^2 * 0.75))\nRange = (0.229167188) * ( 12.7017348 + sqrt(161.334067) )\nRange = (0.229167188) * ( 12.7017348 + 12.7017348 )\nRange = (0.229167188) * ( 12.7017348 + 12.7017348 )\nRange = 0.229167188 * 25.4034696\nRange = 5.82164169 feet\n\nSo the answer is about 5.8 feet
899142
1 feet per second = 0.681818182 miles per hour \n 0.0014 fps=0.00096 mph
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
893501
44.44444 mph.  If the distance of one leg is x miles, then you drove a total of 2x miles in (x/40 + x/50) hours.  Divide the total distance driven by the total time spent driving, and you get 200/9, or 44.444... mph
841492
a)The acceleration a = Force / mass = -6070 / 2600 = -2.33 m/s^2\nInitial velocity u = 34.9 m/s\nThe velocity at any time is given by v = u+a*t = 34.9 - (2.33*1.76) = 30.8 m/s\nb) v^2 - u^2 = 2*a*s, where s is the distance travelled in that time. \ns = ( v^2 -u^2)/2a = 57.8 m\nc) The car stops when u + a*t = 0.\nt = 34.9/2.33 = 15 s
855600
u can also solve this by takin s= -100m\nn by the formula S= ut + 1/2 gt^2\ng = 9.8m/s^2. u = -6m/s .......4 the 2nd part\nand \n\nfind time of the 1st half by the formula \n=> v = u + at....[ where u = 6m/s, v = -6m/s(velocity in oposit direction)]\n\nALSO REMEMBER THAT WEN A TRAJECTORY AGAIN REACHES ITS STARTIN PT.  IT HAS THE SAME MAGNITUDE OF VELOCITY\n\n-ve sign is given to all dat is below the level or in the direction below the startin point\ncombine these 2................ n voila..\nwith time u can master this easy.... n logical method
642567
rise over run tells you how far you move up/down and right/left.\nIf the slope is 4 = 4/1\nthen you can move up 4 and to the right by 1 OR you could move down 4 and to the left by 1.\nIf the slope is -5/8, then you would move up 5 and to the left by 8 OR down 5 and to the right by 8.
671105
almost exactly, assuming someone on earth is doing the measurement. someone moving very fast would get a different result.
241285
For a back row spike I found 26.7-27.6m/s (mean).  Sorry, but that was all I could find!  I was going to do a study on it - but then I switched out of biomechanics into exercise physiology.  You could do it by calculating time elapsed between contacting the ball and the ball reaching the floor.  You would have to know the height at which the ball was contacted and how far it traveled horizontally from the point of contact.  They you find the hypotenuse using Pythagorean theorem and divide this number by the time to get an estimated velocity.
744405
a) (speed is actually closer to 3.97...not a big deal)\n\nb) Conservation of energy:\nThe sum of potential and kinetic energies when the bullet first hits the block and when it is at the top of the swing is the same:\nU(bottom) + K(bottom) = U(top) + K(top)\nNow, the speed at the top is zero, so the ewuation becomes:\nU(bottom) + K(bottom) = U(top)\nSolve for the change in potential energy:\n[U(top) - U(bottom)] = K(bottom)\nThe change in potential energy due to gravity is: (m1+m2)gh\nThe kinetic energy at the bottom is: (1/2)(m1+m2)v²\nSo the equation becomes:\n(m1+m2)gh = (1/2)(m1+m2)v²\nThe (m1+m2) terms cancel:\ngh = (1/2)v²\nSolve for h:\nh = v² / (2g) = 0.8 m
896193
This is an impulse/momentum problem.  Impulse is ∫f*dt[0 - t1] it must equal the change in momentum.  Impulse is given by Favg*t, in your case 0.012*16000 N*sec  Therefore the momentum of the student on impact was 0.012*16000 = m*v.  m is the mass of the student, 63kg.  Therefore his velocity on impact was v =0.012*16000/63.  He fell under acceleration g = 9.8m/sec^2.  Velocity after falling time t is g*t.  Distance traveled under uniform acceleration is h=.5*g*t^.    Use the velocity to get t, then compute h.
823412
Velocity (or speed) defines the time it takes to cover a distance from point a to point b.\n\nV=d/t  >\nV (velocity)\nd (distance in meters)\nt (time in segonds)\nm/s (meters per segonds)\n\nAcceleration is a variation in the velocity, wether it increases OR decreases.\n\nie: gravity on earth has an acceleration of \n-9.98m/s  (notice the minus)  so if an object is falling down on earth it will accelerate towards the ground because the object is subject to that force of gravity, \nif it is thrown up it will slow down as it goes up because the same force of gravity affects the speed of the object, when it has slowed down completely it has reached 0m/s and will start to fall again and accelerate untill it reaches a surface that will stop it from falling further (ground, floor, your hand, etc)\n\nThe formula for acceleration is:\n\na= V1-V2/t \n\nor acceleration = initial velocity-final velocity divided by time.\nif it had an initial velocity that is different from the final velocity there was an acceleration, (or change in velocity).