The vertical and horizontal motions of a projectile are?
821275
4) Independent with one another...\n\nAs the projectile ascend/descend, its velocity has two component.. The vertical and horizontal. Remember that velocity is a vector quantity.\nThe vertical component detemines the maximum height the projectile will attain and also the time of its travel. The horizontal component determines the horizontal distance.
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
239661
The bullets won't be pulled to earth at the same downward _speed_, but they will be pulled by the same downward _force_.  The force that pulls it toward the earth (gravity) would cause both to have the same vertical speed, assuming:\n\n(1) the bullet that was shot from a gun is fired horizontally, and\n(2) these occur in a vaccuum, where the effects of air can be neglected.\n\nNow, to find how long it would take them to hit the ground and how far away the fired bullet would be, you'd need to know how high the bullets are initially and how fast the bullet came out of the gun.  There are a few formulas that you can use:\n\nvf = vi + at\nvf^2 = vi^2 + 2ad\nd = vt\n\nwhere\nvf = final velocity\nvi = initial velocity\na = acceleration due to gravity\nt = time\nd = distance\n\nUsing these in a logical manner can get you what you need.  For example, you can use the second equation to find the vertical velocity of the bullet as it hits the ground.  Then you can take this vertical velocity (vf now) and use the first equation to find how long it took to fall.  Then with this time, you can find out how far the fired bullet went using the third equation.\n\nIt helps if you separate the bullets' velocity into a vertical and horizontal component - i.e. how fast is it moving horizontally and how fast vertically.  These two are independent of each other.  If you keep that in mind that helps in understanding why, in a vaccuum anyway, these two bullets should hit the ground at the same time.
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.
879813
you need to identify that there're two parts of the force, one acting horizontally, the other acting vertically.\nSo:\n1) the vertical part is:\n70sin30\n2) tje horizontal part is:\n70cos30
923444
Short Answer: \n\nvert comp = .161 meters\nhori comp = .936 meters                        \n\nLong Answer:\n\nMb=mass of bullet\nMp= mass of pendulum\nv=velocity of bullet\ng=9.81 m/s/s\nh=vertical displacement\n\n\nBy conversation of momentum:\n\nmomentum of pendulum with bullet=momentum of flying bullet\n\nsqrt(2*g*h)(Mb+Mp)=Mb*v= (28/1000)*230=6.44 \n\nsolving that you can find h=0.161 meters\n\n\nBy the distance formula:\n\nsqrt(2.8^2+0^2)=sqrt( (2.8-0.161)^2 + (horizontal position)^2)=2.8\n\nsolving that you can find horizontal position = 0.936 m
757057
Conservation of momentum\nP = mv before must equal mv after\nmass remains the same\nvelocity is a vecor quantity (magnitude and direction)
733444
horizontal is left to right \nvertical is up and down
1119653
Ok, lets see here...\n\n1st step is to break velocity into horizontal and vertical components (think right triangle here):\n\nvertical velocity = 25.5 * sin(32.1) = 13.55 m/s\nhorizontal velocity = 25.5 * cos(32.1) = 21.60 m/s\n\n\nPart 1:\nFigure out how many seconds before the rock's vertical velocity gets to zero (peak of arc):\ntime up = 13.55 / 9.8 = 1.383 sec\n\nFind the average vertical velocity during that time with linear acceleration from 13.55 m/s to zero: \nVavg up = 13.55 / 2 = 6.775 m/s\n\nFind the distance based on average speed and time:\ndistance up = 6.775 * 1.383 = 9.368 meters above the roof\n\n\nPart 2:\nSo we know that the rock is at 22.368 meters above the ground with zero vertical velocity.  Now, figure out what it's vertical velocity will be when it hits the ground.  \n\nVavg = average vertical velocity on the way down\nVgnd = vertical velocity when it hits\nt = time from peak to hitting ground\n\nWe know:\nVavg = Vgnd / 2\nVgnd = 9.8 * t\nt    = 22.368 / Vavg    = 44.736 / Vgnd\n\nSo...\nVgnd^2 = 9.8 * 44.736\nVgnd = 20.938 m/s\n\nNow convert to a total velocity using the original horizontal component which isn't changed (right triangle again):\n\nVtotal = sqrt(20.938^2 + 21.60^2) = 30.08 m/s\nAngle = inverse tan(20.938/21.60) = 44.11 deg (below the horizontal)\n\n\nPart 3:\nThis is easy.  Horizontal velocity is a constant for the flight (21.60 m/s).  Just multiply by flight time to get distance:\nFrom part2:  time down = 44.736 / Vgnd = 44.736 / 20.938 = 2.137 sec\nflight time = time up + time down = 1.383 + 2.137 = 3.52 sec\nd = 21.60 * 3.52 = 76.02 meters
1065224
Motion of a particle can be defined using the following variables- acceleration, velocity ( initial and final) , time elapsed and displacement. Motion can be translatory only for a particle.It includes linear , circular and projectile.\n\nMotion of a rigid body is defined in terms of translatory motion (motion of c. o. m. of the body) plus rotatory motion, which considers the motion of other particles w.r.t. center of mass.
162313
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
479843
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
869691
a)F = MV=  m1v1 +/-  m2v2  \nusing these we get,  95*4.1 + 85*5.5 (since same direction)=857kgm/s = (95+85)* V. so, V = 4.76m/s\nb) MV = M (6) = (9500+m)*6 = 9500*16 + m*0. \nso, (9500+m) = 9500*16/6= 25333.33 kg\nso, m = 25333.33-9500 =15833.33kg\nc)work done against friction= kinetic energy of bullet\nW = FScos(tetha)  since its horizontal, angle is zero, \nand cos0 = 1, so W = FS\nwe know, coeff of friction = tangent of angle between, the frictional force  and normal reaction...\nso, the \nF = normal reaction=(1.1 +0.015)* 9.81=10.938N\nfrictional force= 0.25 *10.938 = 2.7345N\ndistance is 9.5m...\nso W = 2.7345*9.5 = 25.97 J\nthe kinetic energy of bullet has been transferred to the block...\nso the kinetic energy of bullet is 25.97J\n1/2*(0.015)* v^2 = 25.97\nso, v^2 = 25.97*2/0.015=3463.74\nso, v = 58.85m/s\nhope this helps...:)
925506
You need to consider conservation of momentum and conservation of energy.\n\nConservation of Momentum says that the momentum before tha collision is equal to the momentum after the collision.\n\nm*v1 = (m + M)*v2\n\nwhere m is the mass of the bullet, M is the mass of the block, v1 is the speed of the bullet, and v2 is the speed of the bullet/block system. Therefore,\n\nv2 = m*v1/(m + M)\n\nTo find out how high the bullet/block system rises, consider conservation of energy. The kinetic energy of the bullet/block is converted to potential energy as the system rises so that\n\n0.5*(m + M)*v2^2 = (m + M)*g*h\n\nwhere g is the gravitation acceleration and h is the height to which the bullet/block rises. Solving for h\n\nh = v2^2/(2*g).
891399
assuming constant velocity, the dart is moving horizontally at a rate of 15m/sec at the same time it is accelerating downwards. \n\nso the horizontal distance = 15m/sec * t ot t=2/15 sec\n\nsince acceleration due to gravity is constant, \nvertical distance = (1/2)g(t^2)\n\nso the vertical distance it falls = .088 meters or 8.8 cm\n\nso it falls 8.8cm - 5 cm = 3.8 below the target
1084130
The projectile motion is the motion of a projector starting from origin (x=0 & y=0) with initial velocity (v) & inclination (a) where (a) is the angle between X-axis & (v) so the only accelerations affecting the projector are:\nx''=0 & y''=-g --------> (1) & (2)\n\nTo calculate velocity (x' & y'):\nfrom (1) & (2) we get:       \nx''=dx'/dt=0  \ntherfore: x'=constant=v cos(a) [from initial condition] --------> (3)\n \nfrom (3) we get: y''=dy'/dt=-g\ntherfore: y'=-gt+constant\n\nfrom initial condition:\nat t=0 ----> y'=v sin(a) & constant=0\ntherfore: y'=v sin(a)-gt ------> (4)\n\nThe parametric equation:\nfrom (3): dx/dt=v cos(a)\ntherfore: x=tv cos(a)+constant\n\nfrom initial condition:\nconstant=0\ntherfore: x=tv cos(a) -----> (5)\n\nfrom (4): dy/dt=v sin(a)-gt\ntherfore: y=tv sin(a)-0.5g(t^2)+constant\n\nfrom initial condition:\nt=0 therfore y=0 & constant=0\ntherfore y=tv sin(a)-0.5g(t^2) -----> (6)\n\nfrom (5): t=x/[v cos(a)]          sub in (6)\ntherfore\ny=[x tan(a)]-{ [g(x^2)]/[2(v^2)] }*[1+ tan^2 (a)] \nthis equation is called The Cartizian Equation\n\nTo find The Range ------> sub y=0 in The Cartizian Equation\ntherfor:\nR(range)=[2v sin(a) cos(a)]/g
170531
1) V2^2=V1^2+2d*g ----> 0=47^2+2d*-9.8 ----> d=112.7 m\n2)remeber when mesuring the time we take the velocity on the Y axes and when mesuring range or distance we take velocity on the X axes got it?\nd=V1+0.5*g*t^2 and since the rock is thrawn horizontaly the velocity on the Y axes once it's thrawn is zero, by the way the decplacement would be minus cuz the rock is projecting downwards \n-3.5 m = 0 + 0.5 * -9.8 * t^2 -----> t=0.845 seconds\nd=V*t=47*0.845= 40 m\n3)this is a tough one you'd better draw it :\na.V2^2=V1^2+2d*g ----> 0= (47sin37)^2 + 2d*-9.8 ----> d=40.8m\nb.V2=V1+g*t ---> 0=47sin37+-9.8*t ---> t=2.886 seconds\nrange=V*cos37*t=47cos37=108 m\nc.V2=V1+g*t=47sin37+(-9.8*2)=8... m/sec (velocity on the Y axes)\nvelocity on the X axes is constant and equals=47cos37=37.5m/sec\nthe net velocity^2=37.5^2+8.685^2 ---->velocity of rock after 2 seconds of launch=38.5m/sec