If i know initial position x and y and velocity vectors u and v how can i guess final position of object?
829935
If it's just a moving point, there is no "final" position, but there is a position as a function of time.  Specifically, the position at any time t will be (x + ut, y + vt).  If we are talking about a point moving in a force field, so that u and v are actually functions of x and y, it becomes much more complicated.  You have to treat u as x-dot and v as y-dot ("dot" as in differentiation with respect to time), and then you have to integrate to see if you can get x and y as functions of time.  In this case, one or both coordinates may converge to a finite value as t --> infinity, and then you would have a final position.
654739
The horizontal position of the projectile is described by x(t) = (v_x)*t + h since the horizontal force is zero. v_x is the initial horizontal velocity and h is the initial horizontal position.\nThe vertical net force is -mg, so -mg=ma, thus -g = a so the projectile is accelerated by -g (the acceleration due to gravity at sea level). Integrating, we get y(t) = -(1/2)gt^2 + (v_y)t + k where v_y is the initial vertical velocity and k is the initial vertical position.\nSolving for t in the first equation, we get t(x) = (x - h)/v_x. Since the times in both equations are the same, we plug this value into y(t) to get y(t(x)) = y(x) = -(1/2)g((x - h)/v_x)^2 + (v_y/v_x)(x - h) + k.\nThus, the path of the projectile is a parabola. Setting y=0 will get you the x-intercepts of the parabola, or where the projectile touches the ground.  Note that the projectile starts above ground and time only runs in one direction, so the projectile only touches ground on one root of the equation.
1065479
To find the average velocity, you find the ratio of the position difference to the total time elapsed.  \n\nSo, average velocity \n= [r_vec(2)-r_vec(0)] / 2\n= [(13.8,10.4)-(3.8,0)] / 2\n= (10,10.4) / 2\n= (5,5.2)\n\nTo find the instantaneous velocity at time=a, you find the derivative of r_vec with respect to t for both vector components and then plug in a.\n\nThe derivative of r_vec with respect to t is:\n(5t, 5.2)\n    \nAt t=0, the instantaneous velocity is:\n(0, 5.2)\n\nAt t=1, the instantaneous velocity is:\n(5, 5.2)\n\nAt t=2, the instantaneous velocity is:\n(10, 5.2)\n\nThese units are all in cm, as given in the problem.
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
926868
Solve for momentum.\n\nBefore collision total momentum  = mv\n\nDefine your coordinate system so that the particle was traveling along the x axis in the positive direction, and the collision happens at the origin. Then all of the momentum before the collision is in the x direction. the initial y component of momentum is zero.\n\nparticle 1 goes flying off at v1 at angle a1\n\nparticle 2 goes flying off at v2 at angle a2\n\nThe total momentum afterwards must be equal to the momentum before. So the y component of p1 plus the y component of p2 must equal the y component of the momentum before collision (zero).\n\nSo m*v1*sin(a1) = m*v2*sin(a2)\n\nSimilarly the x components must add up so \n\nm*v1*cos(a1) + m*v2*cos(a2) = mv\n\nNow just solve the first equation for V1 and substitute in the second equation. \n\nThen solve the second equation with substitutions to find the value of v2 in terms of v, and the sin and cos of the two angles. \n\nFinally substitute the value of v2 back into your equation for v1 to find the value of v1.\n\nI leave the simple algebra of solving the equations and the calculation of plugging the numbers into the equations to the student.
614422
Depends what information you already have. The simplest one I can think of is:\n\nVelocity = Acceleration x Time\n\n-Therefore-\n\nAcceleration = Velocity / Time\n\nBut you might have other variables involved here.
855447
if it is equations of motion then \nv=u+ft\ns=ut+1/2ft^2\nv^2=u^2+2fs\nhope this is what you wanted
195576
Y = U + VX^2\nVX^2 = Y - U\nX^2 = (Y-U)/V\nX = sqrt ( (Y-U)/V )
1989
I assume you are talking about vector analysis.  The graphical method would be drawing the vectors head to tail and to scale. and then drawing in the resultant and using your ruler and protractor to measure the displacement and angle.
157080
This is just the regular projectile motion problem (see your text) just with an added initial velocity in the x direction (the velocity of the ship).\n\nLet's take as our positive direction the direction the gun shoots. (That is, the ship is moving in the negative direction.) Thus, the initial x component of velocity of the projectile is the x-component of the guns velocity less the velocity of the ship. (I'm going to use Vo for the initial velocity, since using "not" as a subscript confuses me.)  And I'll use A for alpha to save space.\n\nVxo = Vo cos(A)- v\nVyo = Vo sin(A)\n\nOur equations of motion are\n\nX = Xo + Vox t + 1/2 ax t^2\nY = Yo + Voy t + 1/2 ay t^2\n\nTo find the range, we want when the projectile splashes back down again, i.e. Y = 0.  If we set Xo = 0, then X will be the range, i.e. the X position when Y = 0 again.  And there is no acceleration in the x direction, the y acceleration is -g, the acceleration of gravity.  Plugging all that in, we get\n\nR = Vxo t \nR = (Vo cos(A) - V) t                     (1)\n\n0 = Vyo t - g t^2/2\nVyo t = g t^2/2\nVyo = g t/2\n2 Vyo/g = t\n2 ( Vo sin(A)) / g = t                     (2)\n\nPlugging our expression for t (Eqn. 2) into Eqn. 1 yields\n\nR = (Vo cos(A) - V)2 ( Vo sin(A)) / g \n\nwhich is the formula you gave.\n\nTo find the max range, we can use calculus.  There may be a more straightforward way if you don't know calculus, but this works for me.\n\nThe range formula is a function R of variable A.  To find the max, take the derivative and set it equal to 0.  Thus\n\ndR/dA = 2 Vo/g [sin(A)(-Vo sin(A)) + (Vo cos(A) - V)cos(A)]\n       =2 Vo/g [-Vo sin^2(A) + Vo cos^2(A) - V cos(A)]\n\nusing a trig identity\n\n       =2 Vo/g[Vo cos^2(A) - Vo + Vo cos^2(A) - V cos(A)]\n       =2 Vo/g[2Vo cos^2(A) - V cos(A) - Vo]\n\nWhat we have here is a quadratic equation with the variable cos(A).  When we set it equal to 0, in order to find the minimum, we can use the quadratic formula.\n\ncos(A) = (- (-V) + SQRT[V^2 - 4(2Vo)(-Vo)])/2 (2Vo)\n        = (V + SQRT[V^2 + 8Vo^2])/4Vo
735923
m1*v1 + m2*v2 = (m1+m2)*v3\n\nlets start by calculating in the x direction of V.  The car is traveling purely in the x direction while the van's velocity has no x component (it is traveling south).\n\n1500*25 + 3000*0 = (1500+3000)Vx\n37500 = 4500 Vx\nVx = 8.333\n\nin the y direction:\n\n1500*0 + 3000*(-16.7) = (1500 + 3000)Vy\n50100 = 4500Vy\nVy = -11.1333\n\nSo the resultant vector is traveling south east.  Now using Pythagorean theorem calculate the magnitude:\n\nsqrt(-11.1333^2 + 8.333^2) = 13.9 m/sec\n\nand to get the angle:\n\narctan(8.3333/11.13333) = south 36.8 deg east
889080
1. For two given points P(p1, p2, p3) and Q(q1, q2, q3), the vector v between them is v = (x, y, z) = (q1 - p1, q2 - p2, q3 - p3).\n\nThe magnitude or length of v, denoted by ||v|| is\n||v|| = sqrt[ x^2 + y^2 + z^2 ]\n\nJust do that for every line that you have.\n\n2. This one's a little harder. Let's define the first line as the vector u, and your second line that you're using as the vector v.\n\nYou first need to write each one as a unit vector, which means dividing all of the values through by a value so that the magnitude will be 1. For each of these vectors, this means dividing by the length of the vector.\n\nAs an example, if I wanted to turn w = (1, 1, 1) into a unit vector, I would notice that the magnitude is ||w|| = sqrt[3] and divide all the values through by that to get (1 / sqrt[3], 1 / sqrt[3], 1 / sqrt[3]).\n\nSo, let us say that the new unit vectors are u / ||u|| = (a, b, c) and v / ||v|| = (x, y, z)\n\nWhat these values give you now are called the direction cosines of each vector. If A is the angle between u and the x axis, then a = cos A, and similarly for b with the y-axis and c with the z-axis.\n\nSo, to find the angles between each vector and the x, y, and z axes (also called the direction angles), you take the inverse cosine (cos -1) of the components (values) of the vector.\n\nLet us say that after taking the inverse cosine of u / ||u|| = (a, b, c) = (cos A, cos B, cos C), you get (A, B, C), and the inverse cosine of v / ||v|| gives you (X, Y, Z).\n\nThe angles between the two vectors in the x, y, and z directions are (X - A, Y - B, Z - C).\n\nSo the main steps are finding the magnitude of each vector, normalizing each vector (making them into unit vectors), taking the inverse cosine of each vector, and subtracting the two vectors.\n\nHope that helps.
985616
All are true. For the last one,\nsuppose acceleration = a.\nThen the velocity is at + c\nand the position is at^2/2 + ct + d,\nwhere c and d are constants.
797092
Once an object reaches terminal velocity, acceleration = 0. Thus, you'll be able to determine the terminal velocity by finding an area of the graph where it's a straight line, then taking the slope.
840248
Vertical components\n\nv = u + at\n0 = u + (-32)(2.1s)\nu = 67.2ft/sec\n\nMaximum height the ball reached\n\nv^2 = u^2 +2gs\n0 = (67.2)^2 + 2(-32)(s)\n64s = 4515.84\ns = 70.56\n\ndownward velocity 0.8 seconds after the ball reached the max.height.\n\nv = u +at\nv = o +(32)(0.8)\nv = 25.6 ft/sec\n\ndistance from the maximum height\n\nv^2 = u^2 +2gs\n(25.6)^2 = 0 + 2(32)s\ns = 10.24 ft\n\nHeight of the fence = 70.56 ft - 10.24 ft\n\n\nHeight of the fence is 60.32 ft...\n\nthere ya go...
730085
If x is the angle between two vectors u and v, then:\n\ncos x = (u * v) / ||u|| ||v|| \n\nIf u = (u1, u2) and v = (v1, v2), u * v is the dot product of u and v, equal to u1 v1 + u2 v2, and ||u|| and ||v|| denote the magnitudes of each vector [ ||u|| = sqrt(u1^2 + u2^2), similarly for ||v|| ].
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.