what 2 forces work on a projectile?
147623
After releasing the projectile, \n1. The force due to gravity, towards the center of the earth\n   and\n2. The frictional resistive force due to the medium, for example the air resistance.\n\nFor almost all physics calculations for experiment done on earth, the air resistance is considered negligible.\n\nSo the force due to gravity will be the only force acting on a projectile after it is released.\n\nDanEengineer@yahoo.com
903139
A bullet of course.  A gun uses a small amount of kinetic energy for the trigger hammer to strike the expolsive cap in the shell but thereafter the explosive force is sent into the bullet, with and equal opposite reaction energy dissipated by providing recoil which the gun uses (if automatic) to load the next shell.\n\nI assume the trick in the question is to ignore the equal and opposite reaction rule but not all of the force thus generated is in the form of kinetic energy which relates to bodies in motion.
332766
Force is a vector quantity as it has both magnitude (size) and direction.
733015
a. Applying conservation of momentum\nm1 x V_i + m2 x 0 = (m1+m2) x V_f\nV_f = (m1 x V_i)/(m1+m2)\n\nb. If m1 >> m2 then m1 + m2 -> m1\nTherefore V_f = V_i\n\nc. If m1 << m2 then m1+m2 --> m2\nTherfore V_f = (m1/m2) x V_i\n\nd.m1 x V_i - m2 x V_i = (m1+m2) x V_f\n=> V_f = ((m1-m2)x(V_i))/(m1+m2)\n\ne. m2 = m1, m2 =(1/3)(m1)
655647
Force is a vector and it can be -ve.\n\nThrow a ball up. If you take its displacement upwards to be +ve then the gravity pulling it down is in the opp. direction and hence -ve. Thus the force acting on it is -ve.\n\nIts all a question of taking references.
611013
Oh good grief.  If it's anything like what I've just done it'll include velocity, centrepetal forces, magnetic fields, friction, gravity, forces generally, working out the mass of something and the speed at which it will fall, balances with pivot points, static electricity, AC & DC electricity, transformers, vectors and so on and so on.  It's quite hard but it's maths based so if you like maths, you should be ok.  I wish you luck.
767660
A rocket is subjected to four forces in flight; weight, thrust, and the aerodynamic forces, lift and drag.\n\nat rest I will denote the launch pressure by the symbol ps. The pressure exerts an equal force on all sides of the pressure vessel, the cone of the engine. Because the rocket is free to move along the launch mechisims, the pressure exerts a vertical force given by: \n\nFp = (ps - po) * A \n\nwhere Fp is the pressure force, po is the atmospheric pressure, and A is the cross-sectional area of the rocket tube. The net force acting on the rocket is the difference of the vertical pressure force and the weight W of the rocket: \n\nFnet = (ps - po) * A - W \n\nTo simplify the analysis, we are going to assume that the length of the launch tube is small relative to the length of the rocket. As the rocket moves, the pressure remains fairly constant inside the tube. In reality, there is a slight decrease in pressure as the volume increase, but we are going to neglect that effect to simplify this analysis. With a constant pressure, and constant pressure force, we can use a simple algebraic form of Newton's second law to determine the acceleration a, velocity v and the distance y which the rocket moves along the launch tube: \n\nFnet = m * a \n\nm * a = (ps - po) * A - W \n\nwhere m is the mass of the rocket This equation can also be written as: \n\nW * a = g * ((ps - po) * A - W) \n\nwhere g is the gravitational acceleration, which is equal to 32.2 ft/sec^2 or 9.8 m/sec^2 on the surface of the Earth. The value of g is different on the Moon, and on Mars. The weight W is equal to the mass times the gravitational acceleration. Now divide both sides by the weight to determine the rocket acceleration along the launch tube: \n\na = g * ((ps - po) * A / W - 1) \n\nFor a constant acceleration, the velocity v and distance traveled y are given by: \n\nv = a * t \n\ny = .5 * a * t^2 \n\nwhere t is the time. Using the length of the launch tube L, we can now solve for the lift off time - TLO by using the second equation: \n\nTLO = sqrt (2 * L / a) \n\nand then using this time, we can use the first equation to solve for the velocity at the end of the tube: \n\nv = TLO * a \n\nv = sqrt ( 2 * L * a) \n\nAs the rocket clears the end of the launch tube (t = TLO+), the air inside the rocket, at pressure ps, will rush out the exit and the pressure will equalize with free stream inside the tube. In reality, this will provide an additional amount of thrust T given by: \n\nT = m dot * Ve \n\nwhere m dot is the air mass flow rate out of the tube, Ve is the exit velocity of the air. For an air rocket, with no nozzle and small volume, this additional thrust is small relative to the weight of the rocket, and we will ignore this effect. \n\nThe velocity v at the end of the launch tube is used as the initial velocity in the flight equations to determine the flight trajectory of a rocket.
917347
It sounds like you know what to do.  From your description, it looks like a basic statically determinate problem.\n\nYou have a known applied force P, and two unknown reaction forces, lets say A and B for the top and bottom reactions respectively.  You also have two equations of static equlibrium, specifically the sum of the forces in the "x" direction, and sum of the moments about any point you wish.\n\n(Ok, really you also have the force of gravity and the normal force, but those are the only forces in the "y" direction so that equation won't help you, especially since the mass or weight isn't given).\n\nAssuming to the right is positive, assume A is to the left and B is to the right (this is intuitive, but if directions are incorrectly assumed the answer will just have a negative sign, thats all).\n\n                    _ ------------> P\n                    |   |\n                    |   |\n                    |   |     10.5 m\n                    |   |\n                    |   |\n                    |   |\n                    |   |<----------A\n                    |   |\n                    |   |       1.5 m\n                    |_|--------------> B\n                    \n                 (FBD!)\n\nSum of forces in "x" gives: P - A + B = 0\nSum of moments about B: A(1.5) - P(12) = 0\n\nI took moments about an unknown reaction, B, to eliminate it.  Also, I assumed counter-clockwise to be positive.\n\nSubbing in the value of P = 5.40 N into the second equation:\n\nA = (5.40*12)/1.5 = 43.2 N\n\nThen, put that into the first equation:\n\nB = A - P = 43.2 - 5.40 = 37.8 N\n\nThat should do it.  Both come out positive because their assumed directions were correct, however on the Free-Body-Diagram they are in opposite directions necessary for equilibrium.\n\nCheers,
59102
If the bullet is fired in the horizontel direction, both object will reach ground at the same time.. because they are reaching ground due to vertical gravitational acceleration only which is same for both.\n\nBut if the bullet is fired in the downward direction and at an angle downwards, it will reach grounf first because it is having a initial velocity as well whereas the initial velocity for the other object is zero.. The reverse holds true if the bullet is fired upwards.
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.
840601
(a) Gravity exerts a downward force on the block of 9.8 Newtons/kg * (0.25 kg) = 2.45 Newtons. Work done by gravity = force * distance = 2.45 Newtons * 0.12 m = 0.294 Joules\n\n(b) The spring is a bit tougher since the upward force increases as the spring is compressed. Force = 250 N * (C) where C is the distance the spring has been compressed (in meters). Work is then the integral of Force over distance, so Int(250 * C (dC)) from C = 0 to 0.12 M . The integral is 125 * C^2. Evaluation at 0 gives 0. Evaluation at 0.12 gives 125 (0.12)^2 =  1.8 Joules. So the work done by the spring is -1.8 Joules . Note that the work is negative, because the object is moving against the applied force.\n\n(c) When the block strikes the spring, it will have a kinetic energy of (1/2)mv^2 = (1/2)(0.25kg)v^2 = 0.125v^2.\nThis kinetic energy is absorbed by the spring as it compresses. Note that the work done by gravity is also absorbed by the spring.\nTherefore, we have the following equation:\n0.125v^2 + 0.294 = 1.8\nv^2 = 8(1.8-0.294) = 8(1.506) = 12.048\nSpeed of block just before it hits = SQRT(12.048) = 3.47 meters/sec downward.\n\n(d) I'll leave this as an exercise for you. Just apply the same techniques as above. Note that the block hits the spring with a velocity of 2(3.47) = 6.94 meters per second...
553994
1st - An object in motion/rest will stay in/at motion/rest unless acted upon by another force.\n\nExamples - a ball will keep rolling until its friction with the ground stops it. A rock will stay put unless someone transfers energy to it to move it.\n\n2nd - Force = Mass x Acceleration\n\nExamples - Some car has 5000 kg mass and 10 m/s^2 acceleration. Find its force.\nForce = 5000 x 10\nForce = 50,000 newtons\n\nSome truck has 250,000 newtons of force, and an acceleration of 5 m/s^2. Find its mass.\n250,000 = Mass x 5\nMass = 250,000/5 = 50,000 kg   \n\n3rd - For every action, there is an equal and opposite reaction. \n\nExample - you bounce a ball on the ground. The ball hits the ground, and compresses inward. The ground pushes back up on the ball causing it to land in your hands. You hit a volleyball in PE. The ball pushes against your hands and goes over the net.
730340
Projectile range would be the equivalent of horizontal distance.\nThe forces involved are gravity (acting on the projectile's mass in the vertical direction), and the "push" from the initial thrust.\n\nAs the angle of the firing position increases above the horizontal, more of the initial push is used to combat gravity.\n\nIf the angle is too low, the projectile will meet the ground before it's maximum range (not enough action against gravity).\n\nFor most projectiles, the optimum angle is between 42 and 45 degrees (depending on the strength of the initial impetus). This is because the trajectory approximates a parabola.\n\nWith the given choices, the answer will have to be 45 degrees (answer B)...
1114818
The vectorial sum of the two speeds. If the bullet is fired straight foreward, simply add the speeds. If fired straight back, subtract.\n\nThe setup of your question seems to work with ordinary speed bullets and vehicles, so relativistic effects can be ignored along with  friction.
827544
1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj
901927
Recoil, in common everyday language, is considered the backward kick or force produced by a gun when it is fired. In more precise scientific terms, this force is equal to the derivative of the backward momentum resulting when a gun is fired.\n\nThe backward momentum is equal to the mass of the gun times its reverse velocity. This backward momentum is equal to the sums of the two forward momentums by the law of conservation of momentum and is due to and equal to the projectile's mass times its velocity added to the mass of the escaping gases, burnt propellant, unburnt propellant, and other gases, times their respective velocities. \n\nFor your case, we don't know enough information about the gases and other by-products of the firing. However, if we assume they are negligable, we can focus on the bullets.\n\nConservation of momentum:\nMh * vh1 = Mh * vh2 + Mb * vb\n\nwhere;\nMh is the mass of the helicopter\nvh1 is the initial velocity of the helicopter\nvh2 is the velocity of the helicopter after the firing\nMb is the mass of the bullets\nvb is the velocity of the bullets\n\nDepending on the accuracy of the answer you want, you may want to consider that the mass of the helicopter will change as the bullets leave the gun.\nMh * vh1 = (Mh - Mb) * vh2 + Mb * vb\nhowever, you don't have enough information to solve it in this case. You would need to know the velocity of the helicopter after the firing (vh2). Don't worry about it, I am probably just confusing you.\n\nBack to the question:\nMh * vh1 = Mh * vh2 + Mb * vb\n\nMh * vh1 - Mh * vh2 = Mb * vb\n\nMh * (vh1 -  vh2) = Mb * vb\n\n(vh1 -  vh2) = Mb * vb / Mh
744405
a) (speed is actually closer to 3.97...not a big deal)\n\nb) Conservation of energy:\nThe sum of potential and kinetic energies when the bullet first hits the block and when it is at the top of the swing is the same:\nU(bottom) + K(bottom) = U(top) + K(top)\nNow, the speed at the top is zero, so the ewuation becomes:\nU(bottom) + K(bottom) = U(top)\nSolve for the change in potential energy:\n[U(top) - U(bottom)] = K(bottom)\nThe change in potential energy due to gravity is: (m1+m2)gh\nThe kinetic energy at the bottom is: (1/2)(m1+m2)v²\nSo the equation becomes:\n(m1+m2)gh = (1/2)(m1+m2)v²\nThe (m1+m2) terms cancel:\ngh = (1/2)v²\nSolve for h:\nh = v² / (2g) = 0.8 m