what is a projectile motion?
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heres a good site for u
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Conservation of momentum.  Newton tells us that for every action, there is an equal and opposite reaction, such that in a collision, the net momentum remains the same.  When you realese gas backwards, the momentum imparted on the gas away from the space craft needs to be ofset by imparting momentum upon the astronaut in the direction of the space craft.\n\nWhen you fire a gun, the momentum of the bullet forward is equal to the momentum of he gun backward.  Since the gun is much bigger than the bullet, it doen't move as fast, and the force is spread over a larger asrea (your hands and shoulder).  But the kick backwards is the same sort of principle: the equal and opposite reaction.
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It follows Newton's law that for every action there is an equal and opposite reaction.  It's usually a wheeled vehicle that flings something backward in order to move forward.  The simplest type uses the jet of air escaping from an inflated balloon.
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Wave motion is defined as the movement of a distortion of a material or medium, where the individual parts or elements of the material only move back-and-forth, up-and-down, or in a cyclical pattern.\n\nIt appears as if something is actually moving along the material, but in reality it is just the distortion moving, where one part influences the next.
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Momentum is the tendency of an object in motion to stay in motion. The speed of light is an actual speed.
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Rotational motion is similar to circular motion, except the object involved is a rigid body in which all points rotate around the center of mass of the object and not around a fixed point.\n\nRotational motion can be pure rotational motion or a combination of translation and rotation.\n\nPure rotational motion is circular movement in which all points in the body move in circles, and that the centers of these circles all lie on a line called the axis of rotation. Pure Rotation is caused by an arrangement called a 'force couple'. This is where two equal and opposite forces act on the object from an equal [perpendicular]distance apart.\n\nTranslation and Rotation is caused by a single force which does not pass directly through the center of mass. The center of mass will move in a straight line, meanwhile, all points of the object will rotate about the center of mass.<
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In physics, circular motion is rotation along a circle: a circular path or a circular orbit. The rotation around a fixed axis of a three-dimensional body involves circular motion of its parts. We can talk about circular motion of an object if we ignore its size, so that we have the motion of a point mass in a plane.
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What is the standard way to describe motion ?
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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.
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Only impulse (force x duration, or the quantity of momentum transferred to the pin) is calculable from the data, not force by itself, which is variable in time and not directly meaningful.\n\nIf the moving object has mass M and velocity V, and the object being struck has mass m, then assuming the collision is elastic and all motion is in the forward direction (all these assumptions are at least approximately true for the bowling ball and pin) then the impulse is:\n\nImpulse = 2*M*m*V/(M+m)\n\nEach pound value must be divided by g to get proper units of mass, so this is:\n\n=(g/g^2)*[2*16*3.5/19.5]*V = 5.74*(V/g)\n\nSince:\n\nV/g = 23mph/(32ft/sec^2) = 1.05 seconds \n\nthe result is:\n\n6 pound-seconds\n\nIt's the equivalent of 6 pounds acting for 1 second, or 1 pound acting for 6 seconds, or etc..
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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
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Planetary motion is the motion of the planet itself around its axis\n--> spinning earth\nOrbital motion is the motion of an object around another object in space.\n--> Moon on its path around earth\n--> Earth on its path around the sun
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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.
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The trebuchet is essentially a lever that launches a projectile, with the added complication of a sling.  Therefore, the physics of rotational motion of rigid bodies, the physics of centripetal force (for the sling) and the physics of projectile motion come into the "basic calculations."  As a result, even the basic calculations can be surprisingly complex.\n\nThe first site below just describes the basic physics involved.  The second site actually goes into the calculations.  If you say something more about what you are trying to find, it might be useful.
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muzzle, at which point there is no more force pushing the projectile and the point which air starts to slow the projectile down.\n\nblack powder burns slowly the length of the barrel, pushing the projectile with increasing speed until it leaves the barrel.\n\nSmokeless powder burns rapidly.
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Yes, Kaplan is correct.  While the ball is being thrown it is accelerating (not necessarily at 15 m/s2) up to a velocity of 15 m/s.  Once the ball is released the only force acting on it is gravity at -9.8 m/s2.
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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 :)