A person in a wheelchair approaches a 0.60-m-high ramp. What initial speed would be necessary?
107382
v^2-u^2 = 2gh\nHere, v=1, g=-10m/s/s, h=0.6m\nHence, u^2 = 1+12=13\n=> u = 3.606 m/s
459122
the side length is 2\nso the altitude should be 3^(1/2)\ngood luck
284176
ok well his initial velocity is 0 and the distance is 402m and time is 4.913. you have to find the acceleration. well the only way i can think of doing this  without using these basic equations that are derived from newtons law (i forgot the derivation) is this.     you find the average velocity needed to go 402m in 4.913s. which is just 402m/4.913s = 81.823....\nthis is just the average velocity and because acceleration is constant, this means you can just find the average by doing  Vi + Vf / 2. you have Vi =0    so Vf/2 = 81.823........ so Vf = 163.6474.........m/s\nacceleration is  change in velocity /time. you have change in velocity  (Vf-Vi)   so divide it by time.   \n163.6474-0/4.913   =  33.3 m/s^2\n\n2. so the acceleration due to gravity is 9.8m/s^2. you know the distance is 10m and initial velocity is 0. so Vf = a x t\nso Vf = 9.8t. and Vf+Vi/2 = d/t    same things basically. so Vf/2 = 10/t  which means  Vf x t = 20\nVf = 9.8t\nVf x t = 20\nso then that means that (9.8t) x t = 20\n 9.8t^2 = 20\nt^2 = 2.04....\nt = 1.4285.....  so it takes her that long\nand now you just find Vf by putting this time in the equation  Vf = 9.8t = 14m/s. thats final velocity\n\nb. ok so this question confused me at first but it snot hard. once she hits the water (with speed of 14m/s), she travels for 2m before coming to a stop. so distance is 2m. Vf = 0. Vi = 14m/s and you have to find acceleration. \nso a = (Vf-Vi) x t = -14t\nso you need time now. you do average velocity  again\nVi + Vf / 2 = d/t\n14/2 = 2/t\n7 = 2/t\nt = 0.2857.... s\nthen put it in the other equation and you get a = -14t\nso a = -4m/s^2.\nforce = mass x acceleration\nmass is 61kg so you do 61kg x -4 = -244 N\nthe water exerted 244N against the movement (thts why its negative).\n\nso hope this helped. let me know if you need more help or whatever. laterz
685470
the speed = distance/time
756848
Bob was travelling at 9.5 meters/sec when he started his jump. To gain maximum range he needed to jump at a perfect 45 degree angle. If he had done so, his initial velocity vertically would have been 9.5 /sqrt(2) = 6.72 meters/sec. His maximum height would occur at time t where 6.72-gt = 0.\n9.78t = 6.72\nt = 6.72/9.78 = 0.687 secs\n\nAfter that he would fall for another 0.687 secs till he hit the ground for a total of 2(0.697) = 1.37 secs\n\nIn this time, he would have traveled (6.72)(1.37) meters horizontally or 9.23 meters.\n\nSo his jump of 8.90 meters was 0.33 meters short of perfection.\n\nNote that air resistance (which is pretty important in reality) would have the effect of making the maximum achievable jump shorter than 9.23 meters. So his jump was better than it appeared..
739021
Ok first draw a free body diagram of the puck, at the bottom of the ramp. There's one force pointing down, that's its weight, which is 160/1000*g, where g=9.8ms-2. So the weight is 1.568N.\n\nNow this force can be broken down into 2 components, such that one is pointing parallel to the ramp towards the bottom. This is the force u need to push the puck up the ramp. This force is 1.568N * cos(90deg-22deg) = 0.5874N\n\nF = ma, where F is force, m is mass, a is acceleration.\n\n0.5874 = (160/1000) * a\n\na = 3.67 ms-2.\n\nOk finally, using kinetic motion equation, v^2 = u^2 + 2as, where v is final velocity (in this case 0), u is initial velocity, s is distance.\n\nSo, u^2 = -2as. Don't worry about the negative sign, since acceleration in our case should be negative, since it's pointing towards the bottom,\n\nso, u^2 = -2* -3.67 * 3.8\n\nu = 5.28m/s\n\n\nHope this helps :)
702361
ok biker bike, here we go.\n\nacc.= g = 9.8 m/s2, (any object moving in air will have this acc. or retardation)\n\nb) we apply formula\ns=ut + 1/2at^2\nu=20\nt=4\ns=20*4 + 1/2 *9.8 *4*4\n  =158.4 m\n\nc) v=u + at\n  therefore, t= v-u/a\n                   =50-20/9.8\n                   =3 seconds (approx)
409588
Well I am not sure what "oriaional" means...but if you are asking what the slope of a wheelchair access ramp should be, it should NOT be any steeper than 1:12 or 1 foot long for each inch of verticle rise.  ie; an 8 inch verticle rise should have a ramp that is AT LEAST 8 feet long.\n\nRemember 1:12 is the STEEPEST the ramp should be and may be too steep for some people to use.  \n\nRamps should also have a 5 foot by 5 foot level landing at the top, bottom and at any change of direction and should also have appropriate handrails on both sides of the ramp with 36 inches of clear space between the railings.\n\nI have included a link to the ADA Access Guideline (ADAAG) for ramps from the US Access Board.\n\nRandy G. Wagoner\nWagoner & Associates, Inc.
63054
People inside the plane are still being pulled down by gravity with a force of m*g=9.8*m.  The airplane is also holding the people within it with a force of m*v^2/r=22.22*m.  So you've got a horizontal force and a vertical force.\n\n      22.22*m\n---------------------->\n|\n|9.8*m\n|\nV\n\nPeople are most comfortable when they are accelerated into their seats (not the window or ceiling), so the bank angle will be perpendicular to the sum of those two vectors.  Your result should be about 65 degrees (though the 5 isn't a significant digit-- you should be able to get the final answer on your own).\n\nBut oh those poor people!  They'll be accelerated to almost 3 gs!
704333
let us say that she is walking in the x direction (positive) \n\nThis is a question of momentum... \np = mv   \n(momentum equals mass times velocity)\n\nand we have the conservation of momentum \np_1  +p_2 = p_(total)\nThe sum of the momentums is equal to the total momentum of the system\n\nSo in the begining she is not moving... so the total momentum of the system is zero... it will always be zero.  \n\nShe walks... her momentum is p_1 = 45*1.5 = 67.5 kg*m/s\n\nand we know that p_1 + p_2 = 0\n\nso \np_1 = -p_2 \nor \n67.5 = - m v = - 150 *v\nso the velocity of the plank is -0.45 m/s.  \n\nNow we just sum the velocities to get the total velocity and we see that she is traveling at 1.05 m/s relitive to the ice. \n\nGood question ... i have not thought about a problem like this in a long time
872577
He can accelerate at 0.9g for the whole 50 m. The relevant equation is D=at^2/2, thus t=sqrt(2D/a). Solving, t=sqrt(100/(.9*9.80665)) = 3.366 s, and he reaches a speed of a*t = 29.71 m/s.
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.
441442
you have your units wrong, do you mean 5m/s or 5m/s^2 and you mean 20m from the ground?\nIf acc=5 when height=20 speed= 14.14 in upwards direction (using 2ax=v^2-u^2)\nIf initial speed is therefore -14.14m/s then at t=1 speed=-4.34m/s so it is still going upwards\nt=2 speed= 5.45m/s it is now going down. Both last results worked out using v=u + at\n(Sorry I assumed you meant 5m/s^2 straight up, you need the angle at which it accelerates aswell)
217447
Wight of the top of my head I don't know. How about wheelchairs are us?  Try to contact Goldryder Coaches on the Isle of Wight. They don't sell them, but they would know who does.
1003222
I'm sorry but there are far too many variables to this question.  How gradual is the 90deg turn made?  Which direction is the current flowing?  The person may well be launched but since he is the only person in the boat he, presumably is steering it and as such will be prepared for the 90deg turn.  The boat may also capsize but again, this depends on the points I have made above.
725445
You can't without the radius of the wheel...m/s is a tangential speed.
99451
You could use 2 legs, use a large wedge or just use a ramp type object.