Hard physics question on Projectile motion!!!?
916863
I get that the maximum distance that the ship can fire is 6371m, meaning that any distance greater than 3571m from the western shore should be safe.  It may be tempting to assume that the target ship would be safe within a certain minimum distance, where it would be protected by the mountain peak, but the enemy ship could hit any target within the 3571m limit by angling its guns higher than 45 degrees.\n\nEDIT:\nTo start solving our problem, let's figure out the maximum distance that the enemy ship can fire.  It's not hard, and we know we'll have to figure it out anyway.\n\nIf the ship fires its projectile at exactly 45 degrees, the projectile will make a parabolic arc through the air.  At angles of more than 45 degrees, the projectile will fly higher, but it won't go as far (like a pop-up fly in baseball).  At angles less than 45 degrees, the projectile will fly lower, and hit the water with more horizontal motion (like a line-drive in baseball).  At exactly 45 degrees, it will achieve its maximum horizontal distance.  \n\nIn order to figure out how far the projectile flies, we need to break up its motion into horizontal and vertical components.  Imagine a right triangle with a hypotenuse of 250 units and one angle of 45 degrees.  (The other angle will also be 45 degrees, but that doesn't matter for now.)  The line adjacent to our angle will be as long as the projectile's horizontal momentum.  The line opposite the angle will be as long as the projectile's vertical momentum.  We just need a little bit of trigonometry to figure out these numbers.\n\nsin(a) = opposite / hypotenuse;\nsin(45) = Vv / 250;   (Vv represents the vertical velocity)\nVv = sin(45) * 250 = 177m/s;\n\nSince we're using a right triangle with two 45 degree angles, both sides will be the same length.  If we were using a different angle, we could easily solve for the horizontal momentum using the Pythagorean Theorum.\n\nSo, we know that our projectile had a horizontal velocity of 177m/s and a vertical velocity of 177m/s.  Let's figure out how long it would stay in the air.  We know that it's constantly accelerating towards the earth due to the force of gravity.  Our answer lies in the definition of acceleration: acceleration is a change in velocity over time.  Mathematically, that's (a=ΔV / t).  The "Δ" symbol is the Greek letter delta, often used to represent a change in a value.  \n\nTo figure out how long it stayed up, we only need to worry about the projectile's vertical velocity.  We know that it's initial vertical velocity is 177m/s.  We also know that it's vertical velocity at the top of the arc is 0m/s.  The change between these two points in time is (177m/s - 0m/s =) 177m/s.  The only other thing we need to figure out time is the acceleration due to gravity.  On Earth, it's always 9.81m/s^2. So...\n\na = ΔV / t;\n9.81 = 177 / t;\nt = 177 / 9.81 = 18.04 seconds.  That's how long it takes the projectile to reach the top of it's arc.  It will take equally as long to reach the bottom of it's arc, so the total flight time is 36.09 seconds.  \n\nNow we have a simple distance equation.  Since distance is equal to velocity multiplied by time, the projectile's horizontal distance will be equal to its horizontal velocity multiplied by the amount of time that it stayed in the air.\n\nd = 177 * 36 = 6372m.  (This number is a little higher than the first one, because I rounded a little differently.  It doesn't make a huge difference.)  If we take out the distance to the western shore (2500m + 300M), we get a maximum distance of 3572m from the western shore.  Any ship beyond that distance will be safe.  \n\nAll we need to do now is figure out whether or not the enemy ship can fire at that maximum distance.  If the mountain is too high, the ship will have to increase the angle of its guns, thereby losing range.  We can figure out how high our projectile is going to fly with
1119837
First divide up the initial velocity into its horizontal and vertical components.\nV is the initial velocity of 40 m/s.\nTheta is the launch angle of 37 degrees above the horizon.\nV_x_i and V_y_i are the horizontal and vertical components of the initial velocity respectively.\n\nV_x_i = V * cos(theta) = 40 * cos (37) = 31.95 m/s\nV_y_i = V * sin(theta) = 40 * sin (37) = 24.07 m/s\n\nAlthough the vertical component of the object's velocity will change as the object experiences gravitational acceleration toward the ground, neglecting air resistance, the horizontal component of velocity can be treated as a constant.\n\nV_x = V_x_i = 31.95 m/s\nV_y = V_y_i  - g*t\nwhere t is the time after the object is launched and g is the gravitational acceleration experienced by the object (g = 9.81 m/s^2).\n\nIf the fence which the object hits is 128 meters away (in the +X direction), then the object (traveling at 31.95 m/s in the +X direction) will hit it at about 4.01 seconds after launch.\n\nBy taking this value (4.01 s) and plugging it in for t in the below shown equation to solve for the distance relative to the initial launch height the object has traveled,\nd = V_y_i * t + 1/2 (g)t^2\none can calculate that the object is at a height of about 17.65 meters above its original launch height.\n\nSo I calculate the impact height to be 17.65 meters.
763881
Take the x-axis as the line of action of the moving ball. and suppose the second ball moved at an angle theta wrt to the line of action with certain velocity that u need to find out. Now resolve the velocity components of both balls along x-axis and y-axis before and after collision. Now conserve the momentum along both x-axis and y-axis before and after collision. solve the equations and u will know all the unknowns.\n\nFor the second question find 'e' i.e. co-effecient of restitution and depending on the value it will be elastic or inelastic.
350288
You question is incomplete....or at least requires more information, as it is, it is too general.\n\nWhat are the conditions are are talking about here?\nAre both balls pushed with some constant force?  As they rolled down an incline?  Are they already in motion?\n\nWhat type of surface are they on?  A smooth (frictionless) surface?  A rough surface where friction needs to be taken into account?  Are both balls solid, hollow, one each?\n\nThey are many many variables which need to be taken into account, you cannot just say "big" and "small" and get an accurate answer.\n\nAnd by "greater distance" do you mean translational displacement, or angular displacement?
800064
For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
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
938540
assume shot putter moved the shotput with constant acceleration\nhence we need to find acceration\n\nv^2= u^2 + 2as\n(13)^2= 0 + 2a (2.8)\n169 = 5.6a\na= 30.17857142m/s^2\n\naverage force exerted \n= ma\n= 7 x 30.17857142\n= 211.2N\n\nyey
970380
Let rock be thrown at an angle of θ at a velocity of Vo\n\nThen vertical component = Vo sinθ\nand horizontal component = Vo cosθ\n\nAlso acceleration in horizontal direction = 0 \nand acceleration in vertical direction = -g = -9.8 m/s²\n\nVertical direction\n\na = -9.8\nv = u + at\nSo Vt = Vosinθ - 9.8t\nand s = ut + ½at²\nSo h = Vosinθ*t - 4.9t²\n= 0 when it is at ground level\nie t(Vosinθ - 4.9t) = 0\nSo t = 0 or Vosinθ/4.9 ... ****\nSo it will take Vosinθ/4.9 s to cross the river.\n\nHorizontally\n\nVt = Vocosθ + 0 (as a = 0)\n= Vocosθ\n\ns = ut + ½at²\n= Vocosθ * t\n\nSo 75 = Vocosθ *Vosinθ/4.9\n= Vo².sinθcosθ / 4.9\n= Vo² * 2sinθcosθ / 2*4.9\n= Vo² * sin2θ/9.8 \nThus Vo² * sin2θ = 75*9.8 = 735\nThus Vo² = 735/sin2θ\nNow least value of Vo occurs when Vo² is minimum and this will occur when sin2θ is maximum (=1)\n\nSo Min possible Vo = √735 \n≈ 27.11 m/s (projected at 2θ = 90° ie θ = 45°)\n\nMinimum time = Vo*sin45°/4.9 (from ****)\n≈ 3.91 s
506920
You would agree that, the projectile motion is nothing but a trajectory. It is explained as, the path traced by an object in space.\n\nTherefore when we study the motion of a rocket in space, yes we can call it a projectile.
136454
I will give you a direction to solve the problem...\n\nFirst I will do the problem from a rectilinear formulation(using force, acceleration, velocity) then transfer it to a curvilinear formulation(using Torque, angular accelaration, angular velocity).\n\ngiven F=ma (or F=m(v/t) or more specifically F=m(dv/dt) or even more specifically F=d(mv)/dt).\nWhere F=force, m=mass, v=velocity, and t=time.\n\ngives you Ft=mv (conservation of energy=>conservation of momentum or the impulse-momentum equation).\n\nThis states that it will take a certain amount of force for a certian period of time to produce a certain change in velocity of an object assuming that the mass remains constant.\n\nTransmitting to curvilinear motion force becomes torque, v becomes rv(radius time velocity or w), mass becomes moment of inertia (or I) and t stays the same.
898675
-+he is trying to make you mix up equation and realize what is going on .....\nwhen the canon shoots at an angle the cannon ball will travelin an arc til its reaches the level of the canno and then it will vontinue to move till it hit the ground. always keeping it Ux consant.. so the importn factor in this is the veritcal movement. if up is + and down is - then you through an object upwards and you must find out the time it will take it to reach maximum height and the to -H (SAY HEIGH OF CANNON FROM THE GROUND)\n\n\n\nso -H = Usin(theta)*t - 1/2gt^2 that leads to the \n\n1/2gt^2 -Usin(theta)*t -H = 0 THAT LEADS TO  THE FORM OF \n\nax^2 - bx-c\n\nwotk out the aljebra and you will find 2 solution \nthe negative one is not valid therefore time will be eventually known
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
141479
You should do your own homework problems.  Being told answers will not help you learn.\n\nI'll try to help you understand, though.\n\nThink about this.  The tiger's speed accelerates at a rate of 9.81 m/s^2.  For constant acceleration, distance is related to time by the equation d=t*v_0+(t^2)*a where v_0 is the initial speed IN THE DIRECTION OF ACCELERATION, d is the distance traveled, a is the rate of acceleration and t is the time.  Use that to figure out how much time it takes the tiger to reach the ground.  Then multiply this amount of time by the forward speed.  This tells you how far the tiger has moved forward upon landing.\n\nFor the ball the same formula applies.  It works for the projectile too, but you must decompose the velocity into its vertical and horizontal components (hint:  one will be the sine of the angle it's fired from times the speed and the other will be the cosine of that angle times the speed).  I'm going to leave it up to you to figure out which.  This question obviously has more to do with wanting your homework answered for you than with personal curiosity, because of its specificity, and if you just didn't get the concepts it's ok.  It is not ok to submit your homework on the internet in hopes that others will do it for you.
827118
We can't use conservation of energy because this is an inelastic collision. Conservation of momentum does hold, so\n\nm1*u1 + m2*u2 = (m1+m2)*v\n\nu2 is zero. So \n\nv = m1*u1/(m1+m2) = .24 kg*u1/2.85 kg = .084*u1\n\nThe projectile's incident kinetic energy \nKEp = (1/2)*m1*u1^2 = 0.12*u1^2 joules\nThe final kinetic energy \nKEf = (1/2)*(m1+m2)*v^2 = 1.42*(.084*u1)^2 = .0100*u1^2 joules.\n\nPercentage = 100%*KEf/KEp
915417
Basically, in conservation of energy, you have:\n\nEnergy(initial) = Energy(final).  This becomes\n\nKE(initial) + PE(initial) = KE(final) + PE(final), neglecting air resistance.  When the projectile hits the ground, there is no potential energy because h = 0.  Therefore, the equation simplifies to:\n\nKE(i) + PE(i) = KE(f)\n(1/2)m*v_i^2 + m*g*h = (1/2)m*v_f^2.\n\nWe know that:\n\ng = 9.8 m/s^2\nv_i = 175*sin(45) m/s (because we are only taking the vertical component)\nh = 165 m\n\nThe masses cancel, now solve for v_f:\n\nv_f = sqrt([v_i*sin(45)]^2 + 2gh) = sqrt(175^2 + 2*9.8*165)\n\nSolving, you get v_f = 184 m/s\n\n------------\n\nHope this helps
461302
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.
707702
(2) You need to consider conservation of energy. The energy of the ball at point a is\n\nm*g*Ha + 0.5*m*Va^2\n\nwhich, neglecting losses, is completely converted to kinetic energy at point b. Equating them gives\n\n0.5*m*Vb^2 = m*g*Ha + 0.5*m*Va^2\n\nand solving for Vb\n\nVb = sqrt(2*g*Ha + Va^2)\n\n(3) Same considerations, however when the ball reaches its maximum height (Hm) all of the energy is potential.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hm\n\nSolving for Hm\n\nHm = Ha + 0.5*Va^2/g\n\n(4) If the block is moving at a constant speed up the ramp due to the force F then summing the forces gives\n\nF - m*g*sin(q) - Fr = 0\n\nwhere q is the angle of the ramp, m is the mass of the block, g is the gravitational acceleration, and Fr is the friction force.\n\nSolving for Fr\n\nFr = F - m*g*sin(q)\n\n(5) Again, conservation of energy, but this time at point c the ball has both kinetic and potential energy.\n\nm*g*Ha + 0.5*m*Va^2 = m*g*Hc + 0.5*m*Vc^2\n\nSolcing for Vc\n\nVc = sqrt(2*g*(Ha - Hc) + Va^2)