A level plane releases a bomb with a velocity of 5 m/s from an elevation of 1 km.?
26943
It appears that the question period has expired.  If you have received an answer that meets your needs, please choose one of those as a 'best answer.'  If you haven't received a good answer for your question, you may want to consider the following,\n\n1) Re-post your question.  Newer questions get more activity on Yahoo! Answers than old ones.\n2) If you do re-post your question, consider why it wasn't answered the first time.  Could it be more specific?  Could it be worded better? Were there grammatical or spelling errors? Was it in the best category?\n\nIf it doesn't seem likely that re-posting your question will help you, then here's a listing of my favorite 'answer sites'.  Maybe one of them will help you.\n\nAnswers.com http://www.answers.com/\nBartleby http://www.bartleby.com/\nYahoo Reference http://education.yahoo.com/reference/\nHowStuffWorks http://www.howstuffworks.com/\nWikipedia http://en.wikipedia.org/wiki/Main_Page\n\nSince I really haven't answered your question, it is not necessary to give me any points.  Regards.
949107
It could out fly missiles.  It had a much higher operational ceiling and could fly higher than most surface to air and some are to air missiles.  \n\nI'm not sure on the speed.  At lower altitudes the aircraft operated like a pig.\n\n{edit} To Justin below...  The stealth bomber has never been shot down.  A stealth fighter has, however it is a subsonic aircraft and has very little defense against a well placed missile shot at low altitude.
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
439275
1a)\n\n18 / 7.5 = 36 / 15 = 12 / 5\n12t - 5t^2 = t(12 - 5t) = 0 at 0, 12 / 5\n\nSo at time T = 0 the target is at the origin at the lower left hand corner, and at time T = 12 / 5 = 2.4 the target is at the lower right hand corner (18, 0).\n\n2b)\n\nAt t = 0, rM = (14, 12) is where the missile is fired from.\n\n2a)\n\nThe x coordinate of the missile is a constanr at 14, i.e., the missile is fired straight up.  So it must the target at (14, ?).\n\nThe target moves right according to 7.5t.\n\n14 / 7.5 = 28 / 15 is the time when the target would pass through x = 14.\n\ny = (12t - 5t^2) at time 28 / 15 is\n(12 * 28 / 15) - (5 * 28 * 28 / (15 * 15))\n = (1 / (15 * 15)) (12 * 15 * 28 - 5 * 28 * 28)\n = (5 * 28 / (15 * 15)) (12 * 3 - 28)\n = (5 * 8 * 28) / (15 * 15)\n = (8 * 28) / (3 * 15) = 224 / 45\n\nSo 12 - ct = 224 / 45 at time t = 28 / 15,\nc = (12 - (224 / 45)) / (28 / 15)\n = ((12 * 45) - 224) / (28 * 3)\n = (540 - 224) / (28 * 3)\n = 316 / (28 * 3) = 79 / 21\n\nwith c = 79 / 21\nat time 28 / 15\nthe target is hit at (14, 224 / 45)\n\n3)\n\nI'll let someone else graph something in proportional font ASCII where multiple whitespace is shown collapsed into a single space.
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
440194
Count the seconds from flash to thunder and divide by 5, 1 second 1/5 = of a mile
768444
Plane upwards force > then Earth gravitational pull + mass of the plane+how fare it want to be escape.
224575
its called s5k,since the letter c is pronounced s in russian.it is a small non guided missle of a warhead of 1 kg or so,many of it are carried in a pode under the wings of fighter planes or helecopters.it is used in the role of ground attack.a fairly old missile.
602547
I assume you mean "as viewed by a stationary observer outside the train" - in which case, it will fall straight to the ground, a distance of maybe 1.5 meters.\n\nAnd dude, 1000 mph? That is one helluva fast train.
561411
5 meters from any side
45016
if the rate of rise is 0.917 m/min\nSo it will take 5 / 0.917 = 5.452562 min
869691
a)F = MV=  m1v1 +/-  m2v2  \nusing these we get,  95*4.1 + 85*5.5 (since same direction)=857kgm/s = (95+85)* V. so, V = 4.76m/s\nb) MV = M (6) = (9500+m)*6 = 9500*16 + m*0. \nso, (9500+m) = 9500*16/6= 25333.33 kg\nso, m = 25333.33-9500 =15833.33kg\nc)work done against friction= kinetic energy of bullet\nW = FScos(tetha)  since its horizontal, angle is zero, \nand cos0 = 1, so W = FS\nwe know, coeff of friction = tangent of angle between, the frictional force  and normal reaction...\nso, the \nF = normal reaction=(1.1 +0.015)* 9.81=10.938N\nfrictional force= 0.25 *10.938 = 2.7345N\ndistance is 9.5m...\nso W = 2.7345*9.5 = 25.97 J\nthe kinetic energy of bullet has been transferred to the block...\nso the kinetic energy of bullet is 25.97J\n1/2*(0.015)* v^2 = 25.97\nso, v^2 = 25.97*2/0.015=3463.74\nso, v = 58.85m/s\nhope this helps...:)
754389
Learn to answer these questions yourself!\n\nIf the air is assumed to be frictionless, the vertical and horizontal velocities are independent.  The altitude is given by H = H0 + V0t + 1/2 a t^2.  H0 is +30 meters, V0 is +24.5 m/sec, and a = -9.8 m/sec^2.  If the target is at ground level, you are solving a quadratic for H = 0.\n\nSolving the quadratic gives you the time of flight, t.  Multiply the time of flight by the horizontal velocity (49 m/sec * cos(30°)) and  you get the distance of the target.\n\nOf course, you can just look at the problem and the answers and see if there's just one answer that's anywhere close.  In this case, the initial vertical speed is 24.5 m/sec; it takes 2.5 seconds for the arrow to rise to its peak altitude, and another 2.5 seconds to come back down to the altitude of the bow (5 seconds total).  Falling the final 30 meters would take on the order of a second.  The cosine of 30 degrees is about 0.87, so the horizontal velocity is about 42.5 m/sec; multiply by 6 seconds, and you get about 250 meters.  Ergo, (4) is the only answer that's even close.
641828
500ft\nbut you still have to be careful of whats behind your target.
666681
1/2 * mass * velocity^2.
63054
People inside the plane are still being pulled down by gravity with a force of m*g=9.8*m.  The airplane is also holding the people within it with a force of m*v^2/r=22.22*m.  So you've got a horizontal force and a vertical force.\n\n      22.22*m\n---------------------->\n|\n|9.8*m\n|\nV\n\nPeople are most comfortable when they are accelerated into their seats (not the window or ceiling), so the bank angle will be perpendicular to the sum of those two vectors.  Your result should be about 65 degrees (though the 5 isn't a significant digit-- you should be able to get the final answer on your own).\n\nBut oh those poor people!  They'll be accelerated to almost 3 gs!
1007746
Any radiatioactive isotopes from the blast would have dispersed to near ambient levels in a very short distance.  Certainly in less then 500 miles.