can a moving projectile be in equilibrium?
447650
Equilibrium is defined as the state of undergoing equal forces in all directions (that is mathematically equivalent to zero force).  A moving projectile, with no gravity acting upon it, and no air resistance (i.e. travelling in a vaccuum) would fit that definition.
905071
The trick to solving this problem is to set up your equations properly:\n\nR = Vx * t\n\nH = Vy * t - (1/2)g * t²\n\nWhere R is the projectile's range and H is its initial height, Vy is the projectile's initial vertical velocity and Vx is its horizontal velocity.\n\n[[sorry for the edit, I just noticed the 45° angle; that changes the analysis]]\n\nHere's what we'll do: let's flip the problem around and pretend that instead of landing at a downward angle of 45°, it takes off at an upward angle of 45° with some velocity V, and at some point, t, the projectile has the coordinates (50, 40).\n\nThis introduces two new equations:\n\nVx = V * cos(theta)\nVy = V * sin(theta)\n\nCombining the equations now provides:\n\nR = V * cos(theta) * t\nH = V * sin(theta) * t - (1/2)g * t²\n\nSolving for t in equation 1:\n\nt = R / (V * cos(theta))\n\nSubstituting this into equation 2:\n\nH = V * sin(theta) * R / (V * cos(theta)) - (1/2)g * (R / (V * cos(theta)))²\n\nH = R * tan(theta) - (1/2)g * (R / V)² * (1/ cos(theta))²\n\nWe can now solve for V as:\n\n-H * cos(theta)² + R * tan(theta) =  (1/2)g * (R / V)²\n\n[-H * cos(theta)² + R * tan(theta) * cos(theta)²] =  (1/2)g * (R / V)²\n\nV² = (1/2)g * R² / [-H * cos(theta)² + R * tan(theta) * cos(theta)²]\n\nV = [(1/2)g * R² / [-H * cos(theta)² + R * tan(theta) * cos(theta)²]]^(1/2)\n\nSubstituting the values provides:\n\nV = 49.497 m/s\n\nFinally, plugging this back into the expression for Vx:\n\nVx = V * cos (theta)\n\nVx = 49.497 * cos (45°)\n\nVx = 35 m/s\n\nIf we now, flip our coordinate system back around, we know that the horizontal component of velocity is 35 m/s (which does not change over time). However, this implies that there will be some vertical component of velocity when the projectile is fired. We find this by knowing that the final vertical component Vy equals the final horizontal component (to form the 45° angle):\n\nVx = Vy\n\nAdditionally, the vertical component of velocity changes due to gravity. Thus, the initial vertical velocity component is:\n\nVyo = Vy - g*t = Vx - g*t\n\nWhere t = R/Vx = 50/35 = 1.42 s. Plugging this in, we get:\n\nVyo = 21 m/s (downward)\n\nFINAL RESULTS:\n\nVx = 35m/s\n(Horizontal Velocity)\n\nVy(t) = -21 - 9.8*t\n(Vertical Velocity as a function of time)\n\nH(t) = 40 - 21*t-(1/2) * g *t²  \n(Projectile Height as a function of Time)\n\nR(t) = 35 * t\n(Range as a function of time)\nHope that clears things up a bit! :)\n\nVo = sqrt(Vx² + Vyo²) = 40.82 m/s\n(Launch Velocity)\n\nLaunch Angle = arctan(Vyo/Vx) = -30.96°\n\nTime of Flight = 50/35 = 1.42 seconds\n\n-----------\n\nYes, it's long, I know. But I hope it clears stuff up for you :)
859137
first thing u should do is do draw the path of the projectile,  using only imagination, the ball will definetly fall down as it moves away from the table edge until it finally hits the floor.\n\nprojectile path is always parabolic, and the direction of the velocity vector is tangent to the path of the projectile at any point.\n\nat the initial state, the velocity vector was pointing left or right according to your assumption, but as the parabolic path falls down, the velocity vector points down with it, when the ball hits the floor the velocity vector will be perpendicular to the horizontal in the downward direction.\n\nso at 45 degrees downward, the velocity vector has two components, on in the horizontal V*cos(theta) and another downward V*sin(theta), to get how high the ball at this instant apply the same laws but use V*sin(theta) instead of V ,  and to find how far from the table use V*cos(theta) instead of V.\n\nhope this helped.
994120
The special case of mechanical equilibrium of a stationary object is static equilibrium. A paperweight on a desk would be in static equilibrium. A child sliding down a slide at constant speed would be in mechanical equilibrium, but not in static equilibrium.\n\nIn static equilibrium there is not only a status quo, but there is no motion at all. In dynamic equilibrium there is motion but the overall situation is unchanging.
1161161
The law of conservation of momentum applies in all cases.\nFor a given system, the initial momentum must equal the final momentum.\n\nIn your example, we can define the "System" to be the combination fighter plane + bullet.  If we call this initial momentum P, then at any time after the initial conditions, the sum of the momentums of all parts of our system must add up to P.\nWhen the bullet is fired foward, the fighter plan slows down to compensate and conserve momentum.\n\nMomentum (P) = mass (m) * velocity (v)\nP = mv\n\nThe bullet's mass is very tiny compared to that of the fighter plane, so even a large increase in the bullet's velocity would only mean a small decrease in the resulting velocity of the plane.\n\nIf you call the bullet's mass, m_1 (with final velocity v_2) and the plane's mass m_2 (with final velocity V_2),\nwe know that,\nP = m_1*v_1 + m_2*v_2\nif P_1 goes up, then P_2 must go down by just that amount.\n(delta P_1) = - (delta P_2)\nm_1 * (delta v_1) = -m_1 * (delta v_2)\nsince we assume the mass stays constant,\n(delta v_2) = - (m_1 * delta v_1) / m_2
937015
It varies very widely - it all depends.\n\nSome handgun projectiles are as slow as 600 feet per second.\n\nSome rifle projectiles are as fast as 5000 feet per second.\n\n(Tell him much faster than a race car!)
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.
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
730176
All of these questions can be answered using Newton's 2nd Law:\n\n"The rate of change of momentum of a body is proportional to the resultant force acting on the body and is in the same direction."\n\nThus, if a box has a net force of 0 N, it will not accelerate. This means that its velocity will not change. Note that this DOES NOT necessarily mean that its velocity is 0, just that the velocity is not changing.\n\nIf the box has a net force of 15 N, then it will accelerate in the direction of the applied force. Note that the magnitude of the acceleration is dependant upon the mass of the box.\n\nSo, to answer your two questions:\n\nA) Yes, it will have constant velocity.\n\nB) Yes, the box will accelerate.
863390
I answered your last question and told you about the 45degree angle as being the best.  Here is why:\n\nIn physics dealing with projectile motion an object that is in motion has both a horizontal component and a verticle component.  45 degrees is the optimal angle for projectile motion because it maximises the horizontal component when the verticle component of the motion runs out.  It is called vector motion.  the motion of the marble in the horizontal direction is completely independent of the motion of the marple in the verticle direction.  This is a pretty basic fact.  Your going to get the most distance out of projectiles by shooting them at a 45 degree angle.
660890
If it is balanced, it is in equilibrium by definition.
393015
If a body is fixed at its center of gravity, it is in a state of neutral equilibrium.\n\nTherefore, the whole weight of the body can be assumed to act at the center of gravity.\n\nThe following method is used to find an expression for the C.G of a solid body.\n\nThe solid body is considered to be broken up into elementary particles.\n\nIf m1, m2, m3 etc are the masses of the particles, their weights are m1g,m2g, m3g etc.\n\nThe weights are acting parallel to each other in the downward direction.\n\nIf we assume that the C.G is at the point X,  and x1,x2,x3 etc are the distances of m1,m2,m3, then the  sum of MOMENTS of all these weights about the origin must be equal to the moment of the whole weight of the body about its C.G.\n\nTherefore,   Mg .X =  ( m1g .x1 + m2g .x2+ m3g .x3 + m4g .x4+ ....) where M = sum of all masses.\n\nIn short, the moments of weights of small particles in a solid body about any point is equal to the moment of the whole weight of the body about its C.G.
173849
Not necessarily. Because the collision is elastic, the net kinetic energy of the rubber ball and the sidewalk is conserved. But that doesn't mean that each object on its own has the same initial and final kinetic energy.
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
821414
The bullet's vertical motion is governed by the vertical forces only and not governed by horizontal and vice versa.\nIn each case the bullet's vertical motion is governed by gravity acting on itself. The gravity force gives an acceleration of 9.81 met/sec squared. Both have the same distance to go through and both have a initial vertical velocity of 0. Therefore they both reach at the same time, neglecting air friction.
127576
<equilibrium, state of balance. When a body or a system is in equilibrium, there is no net tendency to change. In mechanics, equilibrium has to do with the forces acting on a body. When no force is acting to make a body move in a line, the body is in translational equilibrium; when no force is acting to make the body turn, the body is in rotational equilibrium. A body in equilibrium at rest is said to be in static equilibrium. However, a state of equilibrium does not mean that no forces act on the body, but only that the forces are balanced. For example, when a lever is being used to hold up a raised object, forces are being exerted downward on each end of the lever and upward on its fulcrum, but the upward and downward forces balance to maintain translational equilibrium, and the clockwise and counterclockwise moments of the forces on either end balance to maintain rotational equilibrium. The stability of a body is a measure of its ability to return to a position of equilibrium after being disturbed. It depends on the shape of the body and the location of its center of gravity (see center of mass). A body with a large flat base and a low center of gravity will be very stable, returning quickly to its position of equilibrium after being tipped. However, a body with a small base and high center of gravity will tend to topple if tipped and is thus less stable than the first body. A body balanced precariously on a point is in unstable equilibrium. Some bodies, such as a ball or a cone lying on its side, do not return to their original position of equilibrium when pushed, assuming instead a new position of equilibrium; these are said to be in neutral equilibrium. In thermodynamics, two bodies placed in contact with each other are said to be in thermal equilibrium when, after a sufficient length of time, their temperatures are equal. Chemical equilibrium refers to reversible chemical reactions in which the reactions involved are occurring in opposite directions at equal rates, so that no net change is observed.
591806
No, this is not perpetual motion, and the ball would not make it up the other side to come out again.\n\nThis is due to the fact that there is air resistance which would slow the ball down, even if it is very aerodynamic.  It would hit terminal velocity rather quickly, and not be able to accelerate any further.  So, in its trip back "up" the other side, it will not have nearly enough velocity to make it all the way up.\n\nThe friction with the air would eventually cause the ball to come to a stop, at a point roughly midway between the ends of the pipe (at the center of gravity).