How long does it take to reach maximum velocity?
342841
It is different for every different circumstance. The key is when the force of gravity pulling you down is exactly equal to the drag on you from the rushing air, you arrive at terminal velocity. If you have a lot of drag (parachute deployed, a lot of loose clothing), it happens pretty quick. If you have low drag (tights, bald) then it takes longer.
103569
Displacement.  If you mean its change in position *divided* by the time it took, that would be its average velocity.
392779
Tthe moving object (mass) travels a distance of one meter in one second and that is an average velocity;that means the speed was not constant as it was moving. So dividing the distance travelled by the period of time traveled would give you only the average velocity.\nThe proof of this was given by a frenchman by the name Paris.\n\n However the final velocity would be twice the average velocity which would be two meters per seconds.dont feel bad I learned that in the 8th grade.
978262
Hi!\nIn a very simplified model it would take about 2.833 seconds. In reality it takes a little longer than that because the air exerts some drag, however he falls. The time depends still on it's weight because the drag depends on the air and shape, while G=m*g depends on it's weight, so the mass remains as a variable. Very light objects such as balloons fall very slow (as you probably know).\n100km/h=27.7 m/sec.    time*acceleration= (desired speed) When there is no initial speed. Thus time= v/accel=2.83sec
940769
ok my friend\ni assume that by now you know that maximum distance is achieved by throwing something at 45 degrees. so start by that and take into consideration that Ux is constant and is Ucos45\nso \nt = 75 / Ucos45\n\n\nnow at 75meters the Height is zero for Uy component\nmeaning that \n\n0 =H = Uyt - 1/2gt^2 =>\n\nusin\n\nUsin45 *  = 1/2 g * 75/Ucos45 ==>\n\nU = sqrt ( 75*9.81) => U = 27.12 m / sec\n\ni used the fact that time is the same and for Ux the distance is 75 but the same time leads to height zero for Uy.\n\n\nnou UX = S/t=>\nt = 75 /Ucso45* \n\nand t=3.91sec
45821
are you sure that you want acceleration instead of velocity??\n\nwell, the acceleration is going to be 0.  since the speed is going to be the same the whole time, there won't be any acceleration.
987748
1.Compute initial velocity V0\nUsing equation S=V0*t-(g*t^2)/2 we find that the initial velocity V0=35/3 + 3g/2\n2.What ever goes up must come down V0*t=-(g*t^2)/2 t=2*V0/g\n3.Total time travel = 2t \nYou to the math\n\nBest\n\nHint:  the arrow will stop midair in about 5.3 seconds
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
797399
velocity=107,300 km/h (or if you prefer 67,062 miles per hour)
1058501
The best thing to do to gain velocity is to work on your legs.  A lot of your velocity comes from the push off on the rubber which uses your legs.  This is simple and you'll see results.
899180
In the direction of the velocity vector. That's the best that can be done, given the limited information you have provided.
751049
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
716223
The change in velocity of an object undergoing uniform acceleration, a, for a given time, t, is given by:\nv = at\nwhere v is the change in velocity of the object.\n\nIf the object is accelerating at a constant rate of 10 m/s^2 for 5 seconds, the change in velocity the object experiences is 50 m/s.\nSince (we assume) the object started from rest, the objects final velocity after 5 seconds is 50 m/s.\n\nTo find the average velocity, divide the total displacement of the object by the total time.\nIn this case, the object has traveled 125 meters in the 5 seconds it has been accelerating (d = 1/2 at^2), so dividing this by 5 seconds and we get an average velocity of 25 m/s.
985453
When it's falling and reaches a velocity where the force from drag is equal to the force from gravity. Simply put, it doesn't speed up any more.
923403
Start off by drawing the picture. You have a right triangle, with a hypotenuse of 25, a base (b), and height (h). \n\nDon't get jumbled down by numbers yet, ignore them. Visualize what you are calculating.\n\nVelocity / speed is the derivitive of position. (v = delta x / delta t). So all you need to do is find an eqation relating the geometry using known and unknown quantities and then differentiate with respect to time. Sounds confusing but all you have to do is....\n\nYou know, a^2+b^2 = c^2 for a right triangle. In our problem,\n\nb^2+h^2 = 25^2\n\nThis is our position equation. Now differentiate with respect to time.  DO NOT forget the CHAIN RULE! <------ VERY important\n\n2*b*db/dt +2*h*dh/dt = 0.\n\nFrom here just use algebra; where,\n\nb=7 ft, and db/dt = 2 ft/sc.\n\nWhich leaves two equations and two unknowns.\n\nThe important thing to remember is to take this step by step and try to visualize what you are calculating. Also, looking at the units of the given quatitities can give you some idea of what you will be required to do. \n\nAlso, just remeber that speed / velocity is the derivative of position with respect to time (the units reflect this).
1008231
Set the equation of motion.\nTaking upward direction as positive\ns=vt-1/2gt^2\nat maximum height s=10,v=0\ng=-g since upward direction is positive\nyou get t= sqrt(2s/g)\n=sqrt(2*10/9.8)\n=1.428 secs
703199
It is an exercise in elementary calculus to show that (ignoring air resistance) the elevation angle for maximum distance is 45 degrees.  The vertical velocity of the egg will be zero at the peak height, and will increase as v = gt^2/2 until it hits the ground.  Energy is conserved, so the vertical velocity when it strikes the ground will match the vertical velocity with which it was launched, and the horizontal velocity will be constant.   This is enough data to solve the whole thing, but there is a shortcut: in the parabolic path at issue, it is always true that the peak height is 1/4 of the horizontal distance.  Hence, the peak height here is 24.225 meters, and that makes it trivial to compute the time and hence the velocity.