hey guys Physics Help (projectile motion)?
826715
bullet drops 0.02 m due to gravity\n\nfree fall distance formula:\n\nd=v1xt + 0.5xgxt^2, v1=0 ( initial vert velocity is 0) and d=0.02 m\n                                   g=9.81\n\nt= (2x0.02/g)^0.5 = 0.0639 s\n\nv=25m/.0639= 391.5 m/s
1119653
Ok, lets see here...\n\n1st step is to break velocity into horizontal and vertical components (think right triangle here):\n\nvertical velocity = 25.5 * sin(32.1) = 13.55 m/s\nhorizontal velocity = 25.5 * cos(32.1) = 21.60 m/s\n\n\nPart 1:\nFigure out how many seconds before the rock's vertical velocity gets to zero (peak of arc):\ntime up = 13.55 / 9.8 = 1.383 sec\n\nFind the average vertical velocity during that time with linear acceleration from 13.55 m/s to zero: \nVavg up = 13.55 / 2 = 6.775 m/s\n\nFind the distance based on average speed and time:\ndistance up = 6.775 * 1.383 = 9.368 meters above the roof\n\n\nPart 2:\nSo we know that the rock is at 22.368 meters above the ground with zero vertical velocity.  Now, figure out what it's vertical velocity will be when it hits the ground.  \n\nVavg = average vertical velocity on the way down\nVgnd = vertical velocity when it hits\nt = time from peak to hitting ground\n\nWe know:\nVavg = Vgnd / 2\nVgnd = 9.8 * t\nt    = 22.368 / Vavg    = 44.736 / Vgnd\n\nSo...\nVgnd^2 = 9.8 * 44.736\nVgnd = 20.938 m/s\n\nNow convert to a total velocity using the original horizontal component which isn't changed (right triangle again):\n\nVtotal = sqrt(20.938^2 + 21.60^2) = 30.08 m/s\nAngle = inverse tan(20.938/21.60) = 44.11 deg (below the horizontal)\n\n\nPart 3:\nThis is easy.  Horizontal velocity is a constant for the flight (21.60 m/s).  Just multiply by flight time to get distance:\nFrom part2:  time down = 44.736 / Vgnd = 44.736 / 20.938 = 2.137 sec\nflight time = time up + time down = 1.383 + 2.137 = 3.52 sec\nd = 21.60 * 3.52 = 76.02 meters
751049
First find Vo:  Vo^2 = Rg/sin(28)  Where R = range & 8 = theta\n\nUsing your numbers, Vo = 41.523 m/s\n\nImpulse = change in momentum ----> Ft = mVo---> F = mVo/t\n\nSo, F = (.0431kg*41.523m/s)/.00438sec = 408.59 N\n\nVoila!
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!
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
479843
I'm assuming it's okay to neglect air resistance in this problem, and that the vertocal velocity component you requested is right before the thingy (I love the word "thingy") hits the ground.\n\nWe have the kinematic equation: S=160+V(initial)t+.5at^2.\na is -32 ft/sec^2, and the initial vertical velocity is 0, so we have S=160-16t^2.  When it hits, S=0, so:\n\n160-16t^2=0 --> 16t^2=160  -->  t^2=10  -->t=10^(1/2)\n\n(Or t = (square root of ten) seconds.  About 3.16 seconds.)\n\nNow that we have t, we just multiply that by the horizontal velocity:\n\n960*10^(1/2), or 960 square root of 10.  (In feet.)  \n\nTherefore the object will land 960*10^(1/2) feet from the base of the building.  Roughly this is about 3035.78 feet.\n\nTo get the final vertical velocity, we use the equation:\n\nV(final)^2=V(initial)^2+2a delta S.  \n\nChange in S in the vertical direction is -160 ft, and a=-32 ft/second squared, and V(initial)=0, so:\n\nV^2=2(-32)(-160)=10240 ft squared per second squared.\n\n--> V=32*10^(1/2), or about 101.19 feet per second.\n\nI hope that's correct.  I typed it with one hand while not really paying attention.  ^_^  From now on, do your own homework.  :P
747360
Well gosh....  \n\nsince the only linear unit mentioned in the question is 18 ft/s  let's assume that the 60 is going to give us acceleration in ft/sec^2 \n\nSo, the position function for this is the second integral of the acceleration function, right? \n\nso...  the first anti-derivative of 60t is  30t^2   which gives the velocity, so at the end of the boost phase, our rocket is moving 30 * 3 * 3 = 270 ft/s \n\nand the second antiderivative is 10t^3  right?  \n\nSo, the position at time T is 10 * 3 * 3 * 3 = 270 feet.  (odd coincidence, but there you are...) \n\nNow, the rocket begins a SECOND acceleration by gravity of -32 ft/s negative 'cause we're accelerating DOWN now... ....   but this time the acceleration is constant,  so \n\nV = gt   and s = 1/2  a  t^2 so, we're slowing from 270 feet per second to zero at the top...  and  t = v/g = 270 / 32 = 8.44 seconds and the height we coast to is the 270 we had at the end of boost plus   1/2  *  32  * 8.44 * 8.44 = 1140 feet.  \n\nThe parachute is going to open after 14 seconds...   so, 14-8.44 is 5.56 seconds of free-fall (and we'll ignore air resistance again) so, we go from zero to something in 5.56 seconds....   and position is 1140 - ( 1/2  * g  * t^2 ) again...   so, the parachute opens at \n\n1140 - (1/2  * 32  * 5.56 * 5.56 )  = 645 feet. \n\nad the velocity is  gt = -32 * 5.56 = -178 feet per second...  \n\nNow, for the NEXT five seconds, we slow under the effect of the parachute   to  -18 ft/sec -- so,  we slow by 160 feet per second in five seconds, so the acceleration of the parachute is (oddly) 32 feet/second squared (huh, how odd is that? )  \n\nSo, the distance traveled during that phase is Vo t + 1/2 at^2  \n\nSo, we have (-160 * 5) + (1/2*32*5*5) =  -800 + 400 = 400 feet.  \n\nand from 400 feet up, we continue down at 18 ft/s for an additional  22 seconds...  \n\nand the rocket lands at t = 22 + 5 +  14 + 3  = 44 seconds \n\nAnd yes, I dropped a bunch of decimal points.  Your original data was only to two significant digits....
152421
Well I wont solve it for you \nbut I will help you. \n\nThese problems always involve the equation \ns=s0+v*t+(at^2)/2\n\nwhere s is displacement\nv is speed\nand a is acceleration\n\nNow you will always have acceleration of gravity g\n\nbut when the freak comes out of the canon, they will have a certain velocity. \n\nThey will be accelerated as the spring releases its energy, and then leave the barrel with a constant velocity, at that angle. \n\nThey will have a horizontal velocity and a vertical velocity. \n\nso you will need two equations\n\nx= Vx *t + .5*Ax*t^2\n\nand y=Vy*t+.5*Ay*t^2\n\nOne for vertical travel, and one for horizontal travel\n\nThe horizontal travel will will assume Ax=0 (not really true, but close enough)\n\nThe vertical travel, assume Ay=-g, acceleration of gravity. \n\nTo find the speed out of the barrel at the start, \n\nremember Hooke's law\n\nF=kx. \n\nSo you know k and you know x, so you know F\n\nYou also know Newton's 2nd law, F=ma.\n\nSo you know a =F/m\n\nNow you have to find the velocity V. \n\nAs the spring releases, it accelerates the freak with acceleration a. The only question is, over what distance is this acceleration acting? \n\nWell one idea is that it acts over the compressed distance. Another idea is that it acts over 2 times the compressed distance, since it will oscillate in approximately simple harmonic motion, and will continue to push that entire distance. I would guess the latter.\n\nSo put this in \n\ns=.5*at^2, \nand compute t\n\nFrom t, you can figure out the speed at launch: \n\nV=a*t\n\nSo you have the initial velocity V, which you can resolve into vertical and horizontal components, and then figure out how long it takes the freak to return back to the ground, and in that time, how far will the freak travel horizontally. \n\ngood luck\n\n- - - - - - - - - -\n\nkirchwey is correct. The same force is not applied through the entire distance that the spring oscillates. The force F=kx is the maximum force, only at the beginning of the freak's acceleration. I was thinking about that so I returned to this problem (it has been a long time since i did these problems). \n\nThe simple way to address this is to assume that all of the potential energy of the spring .5*k*x^2 is imparted to the freak, and turned into the freak's kinetic energy, .5m*v^2. \n\nAs kirchwey points out, there might very well be a new equilibrium position for the spring due to the freak's weight, or at least the component of the freak's weight that is projected along the axis of the tube. If the spring was not at all compressed by anything but the freak's weight, then if the spring was "released" the freak would not go anywhere. What one has to decide is, is the equilibrium point the problem mentions the equilibrium point for an unloaded spring, or the spring with freak compressing it a bit. Whatever you decide will determine how you solve this problem. I would try it both ways and see what the different answers are (they should not be too different because the freak does not weigh that much, relative to the spring force).
802288
the impuls (m*v) is conserved, so the impulse in one direction;\n0.150*20*sin(30deg)\nEquals the impulse in the other direction\n50*v_recoil\nthe recoil velocity is thus 0.03 m/s
680534
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
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
125809
The struggle here is that you provide too little information to recognize this problem as coherent.\n\nWe may assume that the acceleration of gravity is 9.8 m/s/s or 32 ft/sec/sec.\n\nThe velocity(V) of an object is equal to its initial velocity(v) plus its acceleration(a) times time(t). \nV = v + at or V = v - 32t (in feet per second)\n\nYour object will reach maximum height when the velocity is zero or when the initial velocity is equal to 32 ft/sec/sec * time. With your list of times, you'd just multiply by 32 to solve for the initial velocity, if that is what the problem is actually looking for. There are many other permutations of this data, but hopefully that is a start for you.\n\nGood luck.
863271
mass doesn't matter here, just the acceleration\nv=u+at\na=(v-u)/t\na=(9-0)/0.15=60 m/s^2
1053965
Simple parabolic path formula \n\na hit ball is given a velocity v at and angle theta.\n\nAssuming no air resistance the balla travels through the air along its horizontal vector un abstructed, while acted on by gravity in the vertical direction.\n\nthe equation\n\nd=vt + 1/2gt^2 can be solved for d which will be the total vertical distance at time t, where t is the time at which the ball is the highest, lets say about 1/2 of the total time, g is the gravitational constant and v in the velocity in the vertical direction
759333
all object fall 4.9m after 1 sec\n\nX=1/2(9.8m/s^2)(2s)^2\nx=19.6m\n\nx=(1/2)(9.8m/s^2)(3s)^2\nx=44.1m
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.
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.