an object is projected out horizontally, with a velocity of 960 ft/sec., from the top of a building 160 feet high. How far from the base of the building will the object land? (assume a= -32. Find v(suby); v(subx)=960
218060
I'm assuming it's okay to neglect air resistance in this problem, and that the vertocal velocity component you requested is right before the thingy (I love the word "thingy") hits the ground.\n\nWe have the kinematic equation: S=160+V(initial)t+.5at^2.\na is -32 ft/sec^2, and the initial vertical velocity is 0, so we have S=160-16t^2.  When it hits, S=0, so:\n\n160-16t^2=0 --> 16t^2=160  -->  t^2=10  -->t=10^(1/2)\n\n(Or t = (square root of ten) seconds.  About 3.16 seconds.)\n\nNow that we have t, we just multiply that by the horizontal velocity:\n\n960*10^(1/2), or 960 square root of 10.  (In feet.)  \n\nTherefore the object will land 960*10^(1/2) feet from the base of the building.  Roughly this is about 3035.78 feet.\n\nTo get the final vertical velocity, we use the equation:\n\nV(final)^2=V(initial)^2+2a delta S.  \n\nChange in S in the vertical direction is -160 ft, and a=-32 ft/second squared, and V(initial)=0, so:\n\nV^2=2(-32)(-160)=10240 ft squared per second squared.\n\n--> V=32*10^(1/2), or about 101.19 feet per second.\n\nI hope that's correct.  I typed it with one hand while not really paying attention.  ^_^  From now on, do your own homework.  :P
355456
For launch velocity\nv0 = 4.47 m/s,\nlaunch angle\nθ = 66 degrees,\nThe horizontal range is\nR = 1.515173718712374 m.\nThe total time of flight is\nt = 0.83337718278825 s.\nThe peak height is\nh = 0.8507839727702984 m.
454574
This is a simple trig question.\n\nThe horizontal component of the vector can be found using the cosine:\ncos 33.1 = vh / v\ncos 33.1 = vh / 13.6\nvh = 13.6 * cos 33.1\nvh = 11.39\nvh = 11.4 ft sec.
514260
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.
385345
a) There is an equation derived from the motion in x and y that results like:\n\ny = (tanθ )x - g x²/ [2(Vo cosθ)² ] we know y, x and θ we need to solve for Vo:\n\nVo =  (x / cosθ) √[ g / (2 (x tanθ  - y )]\nIf we put  the plane at the initial hit at 1 m, y is 21 m -1m = 20 m\nAnd as we know that x = 130 m and θ= 35°\nSubstituting:\n\nVo = 41.7 m/s <<<<<<<<\n\nb) t = x / ( Vo cosθ )  Substituting: t = 3.81 s <<<<<<<\n\nc) Vfy = Vo senθ - gt   (Vfy is the final velocity of y after the time t)\nSubstituting, its components are:\nVfy = -13.46 m/s      <<<\nVox = Vfx = Vo cosθ = 34.16 m/s   <<<\n\nand its magnitude:\n\nVf = √(Vfx² + Vfy²) = 36.71 m/s <<<<<
554272
If you differntiate the equation you get;\n\ns(t)dt = -32t + 64,\n\nwhen this equation = 0 it gives you the time that it takes the ball to hit it's maximum height. \n\n-32t + 64 = 0  =>  32t = 64  => t = 2.\n\nPlug this into the original equation to get the height;\n\ns(t) = -16*2^2 + 64*2 => -64 + 128 = 64 feet.\n\ni hope that helps.
495208
First of all I assume the height of \nbasket above ground is 2+3.05=5.05\nLet v be the required velocity.\nvertical component=vsin40\n=v*.766\nfrom conservation of energy\nheight gained\n=[v*.766]^2/2g=3.05\nv^2=3.05*2*9.81/.766^2\n=101.986\nv=10.09m/sec
587633
(A) vertical velocity=12.4*sine57.7=10.48m/s\n  Final velocity=0=12.48 - 9.8*t\n  t = 1.07 seconds\n Vertical height = 12.4*1.07-0.5*9.8*1.07^2=5.6m\n Total height=20.1+5.6=25.7m\n Strike velocity= sqrt(2*9.8*25.7)=22.44 m/s\n\n(B) Downward velocity=12.4*sine 57.7=10.48 m/s\n Strike velocity= 10.48+sqrt(2*9.8*20.1) =30.32 m/s
469350
v = v0 - at\n0 ft = 288 ft/sec - 32ft/sec² t\nt =  288 ft/sec / 32ft/sec²\nt= 9 sec\n\ns = 1/2at²\ns = 1/2 (32ft/sec²) (9 sec)²\ns = 1,296 ft\n\nIt's the same time up and down, so 18sec until it hits the ground.
522595
Substitute the h with 180, so 180= -16t^2 + 112t. \n\nThen subtract 180 from both sides of the equation, so 0= -16t^2 +112t -180. \n\nThen factor out a -4 from the equation, so 0= -4(4t^2 -28t +45) \n\nThen divide both sides of the equation by -4, so 0= 4t^2 -28t+45.\n\nThen factor the equation so that 0= (2t-9)(2t-4)\n\nThen use the zero product property so 0= 2t-9 and 0=2t-5.\n\nSolve both and you get t= 5/2 and t=9/2. Remember, the original equation was a quadractic function, so there are two answers.\n\nThis means that the path of the arrow was a parabola, and it first reached a height of 180 ft. at 5/2 seconds, and after reaching its maximum height of 196 ft. at 7/2 seconds, fell to 180 feet again at 9/2 seconds.\n\nHope I got the math right!
525762
x=v(horizontal velocity of water hitting ground)*t( time of travel)\n  = sqrt(2gy) *t; from equation v^2=2gy\nt= sqrt(2h/g); from equation h=.5*g*t^2\nthere you get that equation. I did with kinematics.
439950
V = 2hw², h = 48/w²\nAb = 2w²\nAs = 2wh\nAe = wh\nAl = 2(2wh + wh) = 6wh = 6*48/w\nC = 12*2w² + 5*6*48/w\ndC/dw = 48w - 1,440/w² = 0\n48w = 1,440/w²\nw³ = 30\nw = 3.1072 m\nh = 48/3.1072² \nh = 4.9716 m\nC = $781.44
62152
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
374334
Velocity(at time t) = Velocity (at time = 0) + Acceleration x Time\n\nV(t) = Vi + At\nV(t) = 32 ft/sec + (-32 ft/sec/sec)(t)\nV(1 sec) = 32 - 32 = 0 ft/sec\n\ny(t) = y(initial) + (V x t) + 1/2(A)(t^2)\ny(1 sec) = 0 + (32) +  (-16) = 16 feet\n\nTo hit ground, y = 0, so (V x t) = 1/2(A)(t^2)\n32t = 16t^2\n2 = t, t = 2 seconds\n\ny(max) occurs at zero velocity, which is at t = 1 sec, x = 16 feet
208454
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.
380026
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!
64315
Let V = speed (the magnitude of the initial velocity)\na = angle, t = time, \nVx = horizontal (x) component of velocity\nVy = vertical (y) component of velocity\nAx = acceleration in x direction\nAy = acceleration in y direction\nDx = net distance traveled in x direction\nDy = net distance traveled in y direction\n\nVx = Vcos(a)\nVy = Vsin(a)\n\n  Dx = 1/2 Ax*t^2 + Vx*t\nbut Ax = 0, so Dx = Vx*t, t = Dx/Vx\n\n  Dy = 1/2 Ay*t^2 + Vy*t\n(Ay = -g, the acceleration of gravity, negative for downward)\n -1/2*g*Dx^2/(Vcos(a))^2 + Vsin(a)*Dx/(Vcos(a)) = Dy\nMultiply by V^2, and collect terms with V^2 on the left:\n(Dx*tan(a)-Dy) V^2 = 1/2*g*(Dx)^2/(cos(a))^2\nV = sqrt(1/2*g*(Dx)^2/(cos(a))^2/(Dx*tan(a)-Dy))\nsubstitute a = 30 degrees, Dx = 6.05 m, Dy = 3.05-2.2 = 0.85 m, g = 9.8 m/s^2\nV = 9.51 m/s\nFor a = 60 degrees,\nV = 8.63 m/s