math/physics problem?
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
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
188376
I was a rock star in college physics!   Top of the class!  But I can't remember any of it now.  My teacher always said to start each solution by drawing a picture, then making a list of the varriable you know, and writing down the equations you thik you might need to solve.  It worked for me!\n\nPost your problem, maybe I can help.  Traditional physics problems have not changed much in the last 5 years, maybe I can remember.
797830
Keep in mind these equations\n\nx(t) = x-initial + (v-initial)t + (1/2)(a-gravity)t^2\nv(t) = v-initial + (a-gravity)t\n\n(v-final)^2 = (v-initial)^2 + 2g(x-final - x-initial)\n\na-gravity = -9.8 m/s = -32 ft/s (i think..)\n\nTherefore\n\nx(t) = 0 + (64)t - (16)t^2\nx(t) = -16t^2 + 64t\n\n(v-final)^2 = (v-initial)^2 + 2g(0)\n(v-final)^2 = (v-initial)^2\n\nsince it is going in the opposite direction,\nv-final = -(v-initial)
443144
you should try this site http://www.oberlin.edu/physics/dstyer/SolvingProblems.html
948476
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
695472
Okay,\n\nQuick method:\n\nShe picks up a certain K.E. (kinetic energy) before she hits the water and loses this K.E. 3m under. \n\nWe have KE = Force x distance = mad\n\nso we have m x a1 x d1 = m x a2 x d2\n\nm =65kg, a1= 9.8m/sec, d1=13m, d2= 3m\n\nF(ave)= m x a2= m x a1 x d1/d2 = 2760 1/3 N \n\nThe force exerted on her by the water from when she hits until she stops 3m under will not be constant but the average force stops her in a distance of 3m and has the above value
1135408
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
1135961
in order to solve this problem, you need to work part (b) first before (a)\n\nb): you know gravity (g=32.2 ft/sec) acts opposite the rocket, so calculate, using:\n\nv_final = v_initial - 2*g*y_1\n\nyou know that the final velocity is zero (max height) and you are given the inital velocity. the negative sign is to account for the opposite direction of gravity. move the equation around to solve for y_1 (height from tower to max. height). then add 100 feet (to account for the tower) and you have your answer.\n\na): since you know your maximum height (from part b), substitute that answer for h. [*NOTE: remember to subtract 100 feet from your answer in part b, because t=0 at the top of the tower] You now have a quadratic equation and solve for t using the quadratic formula (you should know what that is). your answer will be the answer!\n\nThis prob is easy only because it worries upto the max height. if we had to worry about the falling down part, well thats another story.\n\n** i already worked out this prob, so if you want to ask more questions/verify your answer, just post an additional response and ill modify my question to answer it. GL!
948754
we juss need to take vector sum of the two forces. assuming the wind is directly in horizontal direction....... sqrt(9.25sq+1.05sq)=9.3N
111044
I'm just doing all your work aren't I?  Its ok I like physics, and its good practice.  I am studying for the MCAT coming up this April (Med School).  This problem unlike the other ones that I did is a little more tricky to me.. Mostly because I don't have my physics textbooks with me.  But it would appear that you need to calculate the acceleration that is exhibited on the foot when it hits the ground.  The way I approached this problem is by using vf^2 = vi^2 + 2ad  and I solved for acceleration.  Knowing that the final velocity is zero I isolated acceleration to get and acceleration of -1024 m/s^2!!! thats huge isn't it.  that is why I'm uncertain about this problem... anyways, multiplying that using the F =ma equation I got a total force of 9216N.  But after inspecting the inital velocity its got to be right because if you convert 6.4 m/s into km/hr you get 23.04 km/hr!!! This person must be falling of a building or something!  So there it is.  Afterall didn't the question start out saying that injury can happen!?
860005
I finished the problem.  The answer is s = 0.48517m.  It was hard, although now that I've worked it, it doesn't seem as difficult anymore.  Had to do some serious thinking though.\n\nThe collar A will tend to slide <down> the bar CD.  The vector CD (from C to D) is [-0.2, -0.3, 0.25] with magnitude 0.43875 and direction cosines [-0.45584, -0.68376, 0.56980].  The weight of collar A, directed downward, is 8g = 78.4N and is represented as [0, -78.4, 0].\n\nWe don't have to worry about the collar being supported by bar CD (the normal force).  The component of the collar directed down the bar is the inner (scalar, dot) product of the weight vector and the direction cosines.  This force is 78.4 x .68376 = 53.60706N.\n\nLet E be the point on the bar CD such that BE is perpendicular to CD.  Then BEA is a right triangle, and by the Pythagorean Theorem, the force along BE is sqrt(150^2 - 53.60706^2) = 140.09384N.  (I'm using numbers stored in the calculator and showing five decimal digits.  All this can be rounded off at the end.)\n\nSince the force triangle BEA is similar to the distance triangle BEA, we just have to get one of the legs to find our distance s = AB.  We will get the distance BE.\n\nTriangle BEA is in the plane defined by the points B, C, and D.  Vector CB = [-0.4, 0.2, 0.15].  The vector cross product CB x CD (see above) is normal to the plane defined by points B, C, and D.  CB x CD = [0.095, 0.07, 0.16].\n\nNow, the cross product CD x (CB x CD) lies in the plane defined by B, C, and D; is perpendicular to CD; and is parallel to the line segment BE.\n\nCD x (CB x CD) = [-0.0655, 0.05575, 0.0145] with magnitude 0.08723 and direction cosines [-0.75091, 0.63914, 0.16623].\n\nNow we can get parametric equations of the lines CEAD and BE, and find the point of intersection E.  Starting with point C and using vector CD, the first line is\n\nx = 0.4 - 0.2t\ny = 0.3 - 0.3t\nz = 0 + 0.25t\n\nStarting from point B and using the direction cosines from vector BE, the second line is\n\nx = 0 + 0.75091u\ny = 0.5 - 0.63914u\nz = 0.15 - 0.16623u\n\nSetting like terms equal to each other, we have three equations in two unknowns.  The three equations are not linearly independent.  Solving the x and y equations for t and u, we get\n\nt = 0.29871 and u = 0.45313\n\n(and these results are confirmed in the z equations).  Plugging these results into the x, y, and z equations, we find that the intersection point E is (0.34026, 0.21039, 0.07468).\n\nWe know the coordinates of point B, so by the Pythagorean Theorem, the distance BE is 0.45313m.\n\nComparing the force and distance vectors in similar triangles BEA, we have\n\ns/150 = 0.45313 / 140.09384\n\nand the length s of the cable AB is 0.48517m.  (Answer)\n\nA difficult, but fun, problem.
976575
Hi Mark, I will answer your question if you answer mine first.\nWhy are you asking this question, it seems like a lot of people in school are trying to take the easy way out of a physics or math problem.\nWill answer if you are not doing said above!\nCheers, Susie
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
826674
use conservation of angular momentum in these kind of problems\nNOT energy conservation\nif you dont know how to use the conservation equation, better read up on examples
179131
Try this:\nhttp://kr.cs.ait.ac.th/~radok/physics/f3.htm
127000
This problem is solved with the application of work and energy principle which states that total work done on a moving particle is equal to the change of its kinematics energy:\nW=T2-T1\nHere the only work done on the particle is dot to gravitation force and is equal to -m*g*h:\n\n-m*10*0.45 = (1/2)*m*(1)^2-(1/2)*m*(v)^2 --->\nm*v^2=(4.5+0.5)*m*2=10*m --->\nv=sqrt(10) m/s\n\nv is equal to sqare root of 10.(approximately 3.162 m/s)