help with projectile motion?
924558
Start off by drawing a right triangle.  The hypotenuse of the triangle is the plane's/rocket's flight path straight into the ground (we'll assume the pilot pulls up after launching his rocket, of course).  The angle is 50 degrees between the flight path and the vertical, so the remaining 40 degrees is the internal angle of the triangle representing the rocket's angle with the ground.  In a right triangle, sine of the angle equals the ratio between the opposite side (the altitude of launch) and the hypoteneuse, and cosine of the angle equals the ratio between the adjacent side (the ground) and the hypoteneuse.  Since you have a speed of 83.3 m/s, you can figure out the vertical and horizontal components of the plane's speed (falling at 53.54 m/s while travelling forward at 63.8 m/s, respectively).  Assuming the rocket followed the exact path the plane would have taken if it crashed into the ground and that it travelled at the same speed of 83.3 m/s, you can use this to figure out what the rocket did -- you know it travelled 750 meters along the horizontal, so divide that by 63.8 m/s and you get 9 seconds of flight time.  You know that if the plane had continued on it's course, therefore, it would have hit the ground in 9 seconds, so it took 9 seconds at 53.54 m/s to go from launch altitude to reach the ground, which means it started at 482 m.
748368
first get the time from the horizontal motion  t = .51 / 1.2 s (uniform motion, no acceleration here)\n\nthen use h = 1/2gt^2 for the vertical motion (with initial velocity zero)
1094911
Make sure your shooting hand is over your elbow.  Here is a good drill.  Assuming your are right handed, stand on a straight line with your right toe lined up on the line.  Act like your shooting and shoot the ball in the air.  It should land on the line, meaning your shot is straight
72876
He means does the angle of launch (theta) affect the range (R), when you launch an object at some constant speed (u)?\nThis is a basic question in applied mathematics (mechanics).\n\nYes it does.\nWrite the equations of motion for the object.\n(Neglect wind resistance, which is secondary.)\nAssuming the launch surface is flat and you did not launch at an elevation, thus y(0)=0,then the components of velocity in x-,y-directions are:\n u_x = u*cos(theta)\n u_y = u*sin(theta)\n\nNoting the acceleration: a_x(t) = 0, a_y(t) = -g (always downwards), then the velocity components are:\n v_x(t) = u*cos(theta)\n v_y(t) = u*sin(theta) - g*t\n\nHence distance traveled is:\n s_x(t) = u*t*cos(theta)\n s_y(t) = u*t*sin(theta) - 1/2 g*t^2\n\nThe object lands at some time t=t' when s_y(t') = 0 again (not the t=0 solution, obviously). Thus:\n 2*u*t'*sin(theta) = g*t'^2\n 2*u*sin(theta)/g  = t'\n...\nand the range R is the s_x traveled at this time t=t' : \n R = s_x(t')\n   = u*t'*cos(theta)\n   = 2*u^2*sin(theta)*cos(theta)/g\n\nNow u,g are constant here, hence range R=R(theta) is purely a function of theta:\n R = (u^2/g) * 2*sin(theta)*cos(theta)\n R = (u^2/g) * sin(2*theta)          ; [a trigonometric equality that sin(2*theta) = 2*sin(theta)*cos(theta)]\n\nIn this case you can show (without even needing differentiation) that R achieves a maximum when:\n sin(2*theta) = 1\n i.e. theta = 45 deg\n which is what you might have suspected.\n\n============================================================\n\n[For the general solution of launching at an elevation, y(0) = y_0 , then you need to rewrite s_y(t), solve as above, using differentiation to show R(theta) achieves a max when R'(theta)=0 for some theta which does not have a neat solution].\n\n[If you want to be a smartass, add the second-order air resistance terms into s_x(t), s_y(t), so you get acceleration due to a viscous force:\n a_x(t) =      -nu * v_x(t)^2\n a_y(t) = -g + -nu * sgn(v_y)*v_y(t)^2\n   [Note we have to use sgn(y_t) to ensure the viscous forces acts against direction of motion. Basically: down when the projectile is rising, and up when it's falling.\nBut you can usually neglect the air resistance component of a_y, since |v_x|>>|v_y| for most if not all of the flight, unless u is small and launch angle theta is large.]\n\nIf you neglect the air resistance component of a_y(t), then you can still get an analytic solution, otherwise I think now you get two difference equations and you have to solve approximately e.g. by Runge-Kutta approximation.\n\nIf you want to be an utter smartass, add in the effect of wind with a vector (w_x,w_y), such that the air resistance force depends on the relative velocity of projectile to wind:\n( v_x(t)-w_x , v_y(t)-w_y ) \n\nAlso, we neglected terrain (mountains etc.), these degenerate into piecewise tests to make sure you always maintain positive altitude s_y(t).\n\nIf you want to get silly, do the ballistic projectile case, where g decreases as a function of altitude, as does nu, and over long distances you have to factor in the earth's curvature, and probably you want to write it in polar, rather than Cartesian, coordinates.\n g(s_y) = g_0  - k_g(s_y(t))   (linearized approximation)\n nu     = nu_0 * exp (-k_nu* (s_y(t)+Earth's radius)) (atmospheric density)\n [but you can linearize this too, for the lower atmosphere anyway: http://www.aerospaceweb.org/design/scripts/atmosphere/ ]\n\nAnd remember this was just for a projectile, where mass=constant. For a rocket, mass is decreasing (non-linearly) over time, and the dm/dt change in mass is proportional to the acceleration force. The direction of the acceleration is of course along the current velocity vector.\nYou may have design limitations imposed on the maximum dm/dt (e.g. the fuel pumps have a maximum throughput). And you could have two- or three-stage rockets.\n\nIsn't it fun how complicated you can make simple things?\n\nHere's a
806029
1st link \n\nb) F = 40 N\n\nc) F = 60 N\n\n2nd link\n\nb) linear velocity = 0.59 m/s\n\nc) Taking g = 9.8 m/s^2\n\nv = 0.83 m/s\n\nd) It would not effect the answer as the expression for v which is sqrt(mu * r * g) is independent of the mass. \n\n3rd link\n\na) Rotate the setup, note the reading on the balance. Find the time required for 1 circuit.\n\nb) No work is done\n\nc) T = 5.48 N\n\nd) 5.9 %\n\ne) If the cord is exactly horizontal, it won't touch the rod and hence cannot be rotated.
507693
I'll get you started...\n\nFirst, determine how long the ball was in the air.  From that, determine the velocity at which the ball was thrown (the rate necessary, factoring in gravity's acceleration, to keep it airborne for that period of time).  Construct a vector using the velocity of the truck vs. the velocity of the ball to determine the angle.\n\nGood luck!
731455
Something that may be fun is a project with model rockets. Give them the necessary information to calculate the height of the rocket, based on the rocket engine and mass of the rocket as homework.\n\nThen at school, fire a few of them and using basic trig calculate the height and see how close they are.\n\nRocket engines are typically rate something like: A8-3 which breaks down as follows:\nA --> 2.50 Newton-seconds of power\n8 --> average thrust (in newtons)\n3 --> seconds from end of thrust to ejection charge (parachute)\n\nThe engines also have a time-thrust curve that shows how the power is delivered over time. Check the following link below for some more info:\n\n------------------------------------------------------------------------\n\nHere is another one:\nNot sure how readily available props are for you but if you can hook up a camera in a room with a strobe light/flash and then set up a collision (air hockey table or pool table come to mind) then you can calculate the velocity of each object, measure its mass and then demonstrate the conservation of momentum using vectors. Because you know the frequency of the strobe, and the distance traveled, you can compute the vector velocity.\n\nOr how about:\nFirst, fire an arrow down a football field. Given the amount of time aloft calculate the initial velocity (ignore air resistance/friction) Then fire an arrow at a bale of hay (extend the bow by the same amount) and measure how far into the bale the arrow went. Using the same initial velocity and distance into bale, calculate the rate of deceleration and coefficient of friction of the hay bale.
981397
This might be one of those questions that you have to answer with just using the formula, i.e. you will not get an actual answer, but an equation that if you knew the mass, length and force you could calculate the height dependant upon the force input, but thats pretty high level stuff, you would need so solve for angular acceleration and momentum, also in this sort of question you would be given either an impact force on a part of the rod (i.e. where the rod was hit) of you would be told the rod has some momentum. Then you could solve.\n\nI suggest you have another look at the question, I think you will find it says the rod is already moving.
956205
First compute the fall time of the keys.  Since you threw them horizontally, the fall is due only to gravity:\n\nd=1/2*g*t^2\n64=.5*9.8*t^2\nt=sqrt(2*64/9.8)\nonce you have t, just multiply the horizontal velocity times the time of flight to get the distance traveled:\n\nd=v*t\nd=8*sqrt(2*64/9.8)\n=29m\n\nIf you didn't understand how to compute the fall time, here's a bit more info:\n\nThe average velocity is\n1/2*g*t\ncompute the fall distance as\n1/2*g*t*t\n\nj
318413
OOO, We did this project in IPC, except our was 20 meters -.-;;. You're ideal angle is 45 degrees. That usually gets you the farthest. It would be pretty smart to use elastic potential energy by using piece of bamboo or just a flexible stick (haha, flexible stick) or just a rubber band. Don't put anything on the egg. Our model was made with a giant rubber band that I found on the floor, and a little seat attached to the crossbow model. If your teacher is strict about a catapult model, your best bet would be to use the bamboo stick and research trebuchets.
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.
739021
Ok first draw a free body diagram of the puck, at the bottom of the ramp. There's one force pointing down, that's its weight, which is 160/1000*g, where g=9.8ms-2. So the weight is 1.568N.\n\nNow this force can be broken down into 2 components, such that one is pointing parallel to the ramp towards the bottom. This is the force u need to push the puck up the ramp. This force is 1.568N * cos(90deg-22deg) = 0.5874N\n\nF = ma, where F is force, m is mass, a is acceleration.\n\n0.5874 = (160/1000) * a\n\na = 3.67 ms-2.\n\nOk finally, using kinetic motion equation, v^2 = u^2 + 2as, where v is final velocity (in this case 0), u is initial velocity, s is distance.\n\nSo, u^2 = -2as. Don't worry about the negative sign, since acceleration in our case should be negative, since it's pointing towards the bottom,\n\nso, u^2 = -2* -3.67 * 3.8\n\nu = 5.28m/s\n\n\nHope this helps :)
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc
942655
Assuming that I understand the geometry correctly, the combine package-rocket will move off the cliff with only a horizontal velocity component. The package will then fall under gravity traversing the height of the cliff (h) in a time T given by\n\nh = 0.5*g*T^2\n\nHowever, the speed of the package-rocket system must be sufficient to cross the river in that time\n\nv2 = L/T\n\nConservation of momentum says that\n\nm1*v1 = (m1 + m2)*v2\n\nwhere m1 is the mass of the rocket, v1 is the speed of the rocket, m2 is the mass of the package, and v2 is the speed of the package-rocket system.\n\nExpressing v2 in terms of v1\n\nv2 = m1*v1/(m1 + m2)\n\nand then expressing the time in terms of v1\n\nT = (m1 + m2)*L/(m1*v1)\n\nsubstituting T in the first expression\n\nh = 0.5*g*(m1 + m2)^2*L^2/(m1*v1)^2\n\nsolving for v1, the speed before impact is given by\n\nv1 = sqrt(0.5*g/h)*(m1 + m2)*L/m1
96240
Use kinematics! Assuming air resistance to be negligible, as it should be for an arrow, the horizontal velocity of a projectile (i.e. an arrow) is constant. Thus, the velocity at which the arrow leaves the bow is constant after it has left the bow. Time the amount of time it takes for the arrow to reach the target after it left the bow. Then, use the fact that v=x/t to find the horizontal velocity of the arrow. Since if the arrow is fired horizontally from the bow, the only velocity imparted to the arrow from the bow is horizontal, the muzzle velocity should be equal to the horizontal velocity that the arrow has in flight. The vertical velocity of the arrow will be constantly changing through out its flight (it will have a constant downwards accelleration due to gravity of 9.81 m/s^2) but, assuming once more that the arrow is horizontal in the bow, there will be no vertical motion in the bow. I suggest using a video camera to time the amount of time it takes for the bow to reach the target after it has left the bow. That way you can use a frame by frame method to get a really precise time of flight. The distance the arrow travels during that time is really easy - it's just the distance between your bow and the target.
918781
Where is your freebody diagram? There are other forces acting on the block than just gravity...\n\nGood Luck
644478
1)Start by modelling the speed of a ball in one direction; take into account friction force (negative drag) that is equivalent to negative acceleration. Loop until speed is equal or less than to zero. Then, stop.\n\n2) Do Exactly the same, now in a 2-D space: model how'd the would move given an initial vector speed.\n\n3) Model a "Hit" on a ball: given an instant (vector) speed plus a "Hit Vector", compute the resultant vector; then use that vector as the initial speed again.\n\n4) develop a "Hit?" function that returns true if the difference between the centers of the balls is less than the diameter, or if the center of the ball is closer than a radius to a wall (the stick can be thought of as another ball)\n5) Every time Hit? is true, recompute vector speed for every involved ball.\nVoila! you have your game. Let me know if it's useful and the results.