3 Physics question with projectile motions! Help?
81906
First Question:\nVertical and horizontal motion are independent from each other.  Use d=(v.i)(t)+(1/2)(a)(t)^2 to find the time that the dart is in the air.  d is distance(0.32m), v.i is the initial velocity(0.00m/s vertically since it was thrown exactly horizontally), a is acceleration (9.8m/s^2: gravity) and t is the time in seconds.  Now that you have the time, use that in the equation x=vt, where x is the horizontal distance (range) that you are asked to solve for, v is the velocity(11.8m/s), and t is the time that you just found.  The answer to the problem is the x variable in this equation.\n\nSecond Question:\nI have attempted this problem for about 40 minutes and I have tried every possible combination of variables and equations, but there is not enough information to solve it.  Therefore, I have to say that it is an impossible problem.  If you figure out how to solve it, please, send me an e-mail at the address I give later and tell me how to do it.\n\nThird Question:\nFirst, find the distance that the basket is from the launch point by subtracting the two.  Then, use (v.f)^2=(v.i)^2+(2)(a)(d), where v.f is the final velocity, v.i is the initial velocity, a is the acceleration (9.8m/s^2), and d is the distance which you just found.  To  get the ball to the basket, it only needs to go the exact distance to the basket, so the final velocity is 0.00m/s.  Use the previous equation to find the intial vertical velocity.  The use v.iy=(v)sin(angle), where v.iy is the initial vertical velocity and v is the initial angular velocity, to find the velocity the ball must be launched at for each angle.\n\nPlease send me an e-mail at bge1234@yahoo.com after you turn in your work and get the solution so that I can see how your teacher did it.  In particular, tell me how to do the second question.  My ways SHOULD work as long as you don't make any calculatory errors.  I hope this helps you.
921378
Projectiles without the presence of air pressure creates a parabolic trajectory (path taken by a projectile).\n\n^_^
846849
1) Yes if you neglect the effect of wind resistance\n2) Yup, because some of its momentum is transferred to the ball (Total remains constant)\n3) Yup, since it was in zero momentum before it was hit and now it has a speed\n4) Ya, all happens
398633
Part 1: Mike\nWork=Force x Distance(moved) x cos (theta) (angle between work and distance)\nF=215\nD=64.7\nTheta=35 degrees\n\nW=215*64.7*cos35\nW=11,394.815 Joules\n\nPart 2: Chris\nF=8N\nD=3.9\nTheta=90 degrees\nW=8*3.9*cos90\nW=0\n\nThats all you get...good luck
49314
Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!
802764
1) KE = ½mv²\n1.1 x 10^9 = ½ x 8 x 10^4 v²\nv² = 2.75 x 10^4\nv ≈ 165.8 m/s\n\n2) another physics question: what is the speed of a 0.145kg baseball if its kinetic energy is 109J?\n\nDo as above\n\nKE = ½mv²\n109 = ½ x 0.145 x v²\nv² ≈ 1503.4\nv ≈ 38.8 m/s\n\n3) ANOTHER QUESTION: TWO 3.0 bullets are fired with speeds of 40.0 m/s and 80.0 m/s, respectively. what are their kinetic energies? which bullet has more kinetic energy? what is the ratio of their kinetic energies?\n\nAssuming 3.0 g bullets as they would NOT be 3.0 kg bullets\n\nSo 3 g = 0.003 kg\n\na. v = 40 m/s\nKE = ½mv²\n= ½ x 0.003 x 40²\n= 2.4 J\n\nb. v = 80 m/s\nKE = ½mv²\n= ½ x 0.003 x 80²\n= 9.6 J\n\nratio of energies (slower to faster) = 2.4:9.6 = 1:4 = (ratio of velocities)²\n\n4) another physics question: a car has a kinetic energy of 4.32 x 10^5J when traveling at a speed of 23 m/s. What is it mass?\n\nKE = ½mv²\n4.32 x 10^5 = ½m x 23²\nm ≈ 1633 kg
945234
[Edit -- two things.  #1 Oops, I made a mistake.  When I worked this, I used a mass of 4 kg instead of 0.4.  Now I'll go back through my answer and make the correction.\n\n[#2  There are two different methods here.  Look at both of them.  The <second> way is <much better> for this problem.  I think you'll see that.]\n\nSnow White has an answer, but she doesn't say how she got it.\n\nLet's start by getting the net force F = F1 + F2 = -0.6Ni + Nj with magnitude sqrt(1.36)N = 1.166 N and acceleration F/m = 2.915476 m/s^2.\n\nNow we have two ways to proceed.  This is linear motion with constant acceleration, and we know the direction ... tan theta = -1 / 0.6 = -5/3, so the angle theta in the second quadrant is 59 degrees north of west.\n\nAfter 1.6 seconds, the velocity is acceleration x time = 2.915476 x 1.6 = 4.66476 m/s in the direction indicated above.\n\nThe displacement s after 1.6 seconds is s = vt/2 or s = 1/2 at^2 = 4.66476 x 1.6 / 2 = 3.7318 m.\n\nTo get the position, we could use sines and cosines, but notice that tan theta above is -5/3, so the hypoteneuse of that triangle is sqrt(25+9) = sqrt(34).  Using that geometry, the position after 1.6 seconds is\n\nx = -3/sqrt(34) x 3.7318 = -1.92 m\n\ny = 5/sqrt(34) x 3.7318 = 3.2 m\n\nThat's your answer: (-1.92, 3.2) with velocity 4.66476 m/s and direction 59 degrees north of west.\n\n\nThere's another way to do it that may be better, especially when you're using vectors.  Do it by components, and keep the components separate.\n\nThe net force is F = F1 + F2 = -0.6Ni + Nj.\n\nThe constant acceleration a = F/m = -1.5i + 2.5j m/s^2\n\nThe velocity after 1.6 seconds is v = at = -2.4i + 4j m/s\n\nThe position after 1.6 seconds is s = vt/2 = -1.92i + 3.2j\n\nand this is the same answer we got by doing it the other way.  (You can, of course, get all the magnitudes and angles.)\n\nLooks to me that it's easier doing it by components.\n\nHope this is the same answer Snow White got.  This way, at least, you have two ways of doing it.
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.
438060
Without regard to air resistance you solve the problem as such.\n\nv(f) = final velocity\nv(i) = initial velocity\na = acceleration\nt = time\nx = displacement (distance)\n\nWhen you throw the ball up vertically, you know several things:\n1) v(i) = 30 m/s\n2) v(f) = 0 m/s\n3) a = -9.8 m/s^2 (this is because acceleration due to gravity goes in the opposite direction of the object)\n\nSolve for x with this equation:\nv(f)^2 = v(i)^2 + 2ax\n0^2 = (30 m/s)^2 + 2(-9.8 m/s^2)x\n0 = 900 m^2/s^2 - (19.6 m/s^2)x\n(19.6 m/s^2)x = 900 m^2/s^2\nx = (900 m^2/s^2)/(19.6 m/s^2)\nx = 45.9 m (rounded off)\n\nNow you know how high the ball went.  The time it takes to reach this is equal to the time it takes to get back down (assuming that it is the same distance up and down).\n\nTo solve for t, use this equation:\n\nv(f) = v(i) + at\n\n0 = 30 + (-9.8)t\n9.8t = 30\nt = 30/9.8\nt = 3.06 s  this is going up to max height.\n\nto check if it will be the same time going down, use this equation:\n\nx = v(i)t + (1/2)at^2\nv(i) = 0 (because when you reached max height, your velocity will be zero)\na = 9.8 because your ball is travelling in the direction of the acceleration.\n45.9 = 0 + (1/2)(9.8)t^2\nt^2 = 45.9/4.9 = 9.367\nt = 3.06 s\n\ndone
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
809913
1. Momentum=Massxvelocity=3600\n2.3600=.2KgxVelocity=18000\nThats all i know
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.
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If everything were lined up exactly inline and marble 1 hits 2 which hits 3, then both marbles 2 and 3 would move together at the same rate as marble 1.    (1~~>2~~>3)   You have probably seen that demonstrated with the newton's pendulum toy that has 5 chrome balls hanging by strings.   When you pull 1 of the balls back and let it swing into the other 4, 1 ball swings off the other end.  2 balls cause 2 to swing off the other end.\n\nIn your experiment, the 1st marble will hit the other 2 and release all of it's energy.  They will then go at the same rate as the first marble.\n\nV1 = 0\nV2 = 15.5 in a direction 60 degrees off to left.\nV3 = 15.5 in a direction 60 degrees off to the right.
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
170531
1) V2^2=V1^2+2d*g ----> 0=47^2+2d*-9.8 ----> d=112.7 m\n2)remeber when mesuring the time we take the velocity on the Y axes and when mesuring range or distance we take velocity on the X axes got it?\nd=V1+0.5*g*t^2 and since the rock is thrawn horizontaly the velocity on the Y axes once it's thrawn is zero, by the way the decplacement would be minus cuz the rock is projecting downwards \n-3.5 m = 0 + 0.5 * -9.8 * t^2 -----> t=0.845 seconds\nd=V*t=47*0.845= 40 m\n3)this is a tough one you'd better draw it :\na.V2^2=V1^2+2d*g ----> 0= (47sin37)^2 + 2d*-9.8 ----> d=40.8m\nb.V2=V1+g*t ---> 0=47sin37+-9.8*t ---> t=2.886 seconds\nrange=V*cos37*t=47cos37=108 m\nc.V2=V1+g*t=47sin37+(-9.8*2)=8... m/sec (velocity on the Y axes)\nvelocity on the X axes is constant and equals=47cos37=37.5m/sec\nthe net velocity^2=37.5^2+8.685^2 ---->velocity of rock after 2 seconds of launch=38.5m/sec
1120781
do your own homework\n\nits probably 0.26 kg M/s ^2 since its a velocity\n\nyou also probably need to convert it to rad/s whatever that is
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Let the two case be identified as having  moment-of-inertia I1 and I2. \nTorque T on to the each of the two bodies produces angular accelerations (alfa1 and alfa2) given  by the relation T= I x alpha \n\nalfa1 = T/ I1 and alfa2 = T/I2\n\nStarting form rest(wi =0), let the time taken to reach final angular velocity wf be t1\nusing:  wf =- wi + alfa x t  \n \nwf/alfa1 = t1 and wf/alfa2 = t2 \nleading to \n\nt2  = alfa1x t1/alfa2 \n\n     = I2 xt1/I1\n     = (I2/I1) x t1\n     =. ......Ans \n(Hoping you will put the values and get the answer you want)