This is a hypothetical physics question.\nA helicopter takes off vertically from ground and stays at some height for 12 hours.It lands back after 12 hours. Will it land at the same location from where it took off OR will it land on the other side of the globe?
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I am assuming that they are assuming that there is absolutely no air acting on the helocopter except to keep it up AND that the helocopter pilot can somehow go directly straight up and down perfectly without drifting one way or another due to bad control.\n\nOK there is gravity acting on the helocopter, and it's initial velocity is the velocity of the face of the earth. So again you have to make an assumption: Is the helocopter on the equator, one of the poles, or somewhere in between? If the helocopter is on the North or South pole, then it will land exactly where it took off. However, if the helocopter takes off from a point on the equator, it will remain a the same lateral velocity as the surface of the earth, but it's distance from the center of the earth (raduis of the circle it is traveling on INCREASES. This means that the earth will continue on to the EAST, so the helocopter will land somewhere slightly to the WEST of where it started!\n\nNow if the chopper is anywhere between the equator and the pole, the distance it travels to the West depends on how far from the pole it is. We know it will not move from it's spot on the pole, and we know it will move the furthest at the equator. So the distance would be somewhere between those two. But there is MORE... If you are not on the equator, you have a north or south component to your velocity because of the fact that you are tuning to the side in a circle away from the equator. This makes sense when you consider that you are rotating in place when you are on the pole. So in addition to landing slightly west, you would also be flung slightly toward the equator for and starting spot between a pole or the equator. So the answer is that if you are in the northern hemisphere, you will drift slightly SOUTHWEST, and if you take of in the southern hemisphere, you will drift slightly NORTHWEST.\n\nThe teacher may as you, "how can we tell how far to the west we will drift?" Well remember the difference in radius between the earth surface and the height of the chopper? Well the higher you go in that 12 hour span, the slower your angular velocity, meaning the longer it takes you to travel along the circle you are tracing. Obviously if you climb up 1 foot, you will not be very far from where you started 12 hours later, however, if you use you imagination, and rise up to 1,000,000,000 miles in a second(assuming you ignore gravity from the sun and other planets), and drop back down to earth in the last second, you would be very close to the other side of the world. So somewhere between those two extremes is exactly how far you will drift.\n\nDon't be tempted to think that the helocopter will be in space somewhere due to the idea that the earth is moving around the sun because that becomes negligible when you consider the strength of gravity at a low altitude.\n\nIf your teacher wants you to assume the atmosphere is acting on the helocopter then you need a meteorologist to determine where the helocopter lands.
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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
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OK, relativity DOES NOT come into play here, because the velocities involved are not remotely high enough.\n\nYou are talking, it seems, about an object being projected up into the air and falling back to earth.  This must occur below escape velocity or it would leave the earth.  Relativistic speeds are far, far, far, far higher than that.  So, your question is in fact not related to relativity in any way, as much as you seem to think it is. SO I will answer your question 'normally'.\n\nWhen the object is projected upwards, it possesses a certain amount of kinetic energy whcih, as it rises, it converted into potential energy which increases with height.  When the KE finally runs out, the object will peak and as gravity, which exists by the way, will pull it down the PE will be converted back into KE and thus motion, in a downward direction in this case.\n\nAll I can say after this is to stand under it and maybe after it hits, you'll understand a little better.
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Our atmosphere is considered the upper layer of our planet, meaning that it rotates with the planet.  So due to friction and gravity you will stay in the smae position relative to the ground below you basically rotating with the Earth, unless it is windy.
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None. It takes no power to maintain that speed (on level ground, no friction). The object has kinetic energy (0.5 * m * v^2). Power was required to put it in motion and power can be recovered by decelerating it. If you do the math, the energy is 3551.86 ft-lb, equivalent to 1 horsepower for 6.458 seconds. Answer 1 neglected the fact that weight must be converted to mass by dividing by g, and energy is proportional to v^2.
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Do you already know where the CM of the hemisphere is?  Because I don't and it would take me a while to work it out.  Anyway, it would be on the vertical radius, so the horizontal position of the CM of the composite body is on the radius drawn to the particle, distant 0.2*r/4.2 from the centre.  If the height of the CM of the hemisphere above the centre is h, then the height of the CM of the composite body is\n4*h/4.2
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****Edit based on further details from asker *****\nSailing around the world is sailing within the sperhical confines of our planet.  If there were no athmosphere, or gravity, then we would simply fall off the earth.  However, as gravity is pushing us down, and is affecting the winds, ect within the athmosphere, we are bound to the surface of the water.  \n\nEven though a sphere is included in the define of spheroid, it is a type of spheroid, not the inclusive answer.  The fact that gravity pushes down on us from all around, mainly because the earth is spinning so fast, and other more complicated reasons; the sphereoid holds whatever is on it's surface down to adhere to the confines and shape of such a surface.\n\nEven if the world were flat, or a flattned sphereoid, we woudln't fall off, instead, we'd loop around to the otherside becasue gravity wouldn't allow us to actually 'drop' off the side of the cliff.  The inverse gravity on the other side of the plane would pull us to it's confine, or surface.\n\nSo, there's your answer.  Sorry i didn't get it more clear the first time.\n\nMathamatically you can say that gravity pushes us from all around using a tangent direction.  That means that no matter the point you are on this earth, gravity is pushing down on you within the same confines as any other like region.  As to falling off the earth, if gravity didn't exist, we wouldn't 'fall' exactly; we would more likely be thrown or float off the earth (thrown if the world is spinning).\n\n****End Edit.******\n\nThe world indeed is a sphere, but it is not a perfect sphere, due to the rotation of the earth around it's axis, as well as the rotation of the earth around the sun (central axis), the spinning and speed of the planet make it more of a spehere that jiggles, sometimes squeezed, and sometimes elongated.  But, not by much :) Just a bit :)
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I think you can use a simpler equation than the one for distance.\nWe don't need to find distance, just time.\n\nv = u + a.t\nv= 0, u= 64 f/s, a= -32f/s^2\nsolve for t. \nt = 2 s\nThats the time it takes to go up just before it starts to fall again.\nThe time taken to fall is going to be the same: another 2 s\n\ntotal time: 4 s
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To figure out how long it will take the object to hit the ground, we can use the distance equation: d=1/2at^2 ---> t=√2d/a\nThe distance is 100m, and the acceleration is 9.8m/s^2: \nt=√2(100)/9.8 --->t=4.5s\nNow that we have the time it's in the air, we can figure out the horizontal distance by using the velocity equation: \nv=d/t --->d=vt\nThe velocity is 20m/s, and the time is 4.5s, so: \nd=(20)(4.5) ----> d=90m\n\nAs for question C, the objects horizontal velocity (ignoring air resistance) will stay the same througout its flight, so it hits the ground at the same horizontal speed it had when it started: 20m/s\n\nAnd to figure out question D, just remember that velocity will equal acceleration times time. So: \nv=at ----> v=(9.8)(4.5) --->v=44.1m/s\n\nHope this helps!
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Have people never seen a real helicopter close by.... of course the rotors appear to go backwards... It is an optical illusion but it is exactly the same as a spinning bike wheel. When it reaches a certain speed the eye See's a part of it slow down and then it appear to go slowly backwards...... until it's just a blur again....\n\nI will go look for what the term for this is called and will be back with an edit shortly ;o)\n\nOK Back. I found this....\nImagine a clock with just a minute hand. If we were to look at the clock every 59 minutes it will appear to have gone backwards by one minute each time and if we look every sixty minutes it will appear not to have moved at all! So to accurately observe the minute hand in motion we must look at the clock more frequently than every sixty minutes, in fact we must look at it every thirty minutes. This is known as Nyquist's theorem.\n\nGraham
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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
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They're already flying East.  That's why they can't stop.  Let's just say for the sake of argument that anything attached to the surface of the Earth stops instantly.  We do that and there are still interesting things to talk about.\n       Anything not attached in that case would go flying East.  But how fast it went would depend on how close it is to the equator or the poles.  If you were on the south pole or north pole, you would just stop slowly rotating.  If you were near one of the poles, you would fly east, but not so fast.  If you were in the continental U.S., you would go east somewhere around 600 - 700 mph depending on your latitude.  If you were at the equator, you would fly east at around 1000mph.
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Definitely false - the pull of gravity is a constant 9.8 m/s^2.  The only thing that would alter them landing at the same time is wind and the air resistance.  Also, as someone else pointeed out, if someone was standing underneath the path of one of the bowling balls (but that's unlikely).
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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.
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Very impractical for at least two reasons:\n\n1) It would have to be an unbelievably powerful magnetic field to generate any significant lift. The Earth's magnetic field is very weak.\n\n2) As soon as it left the ground the vehicle would be trying to flip over to an attitude where the Earth's magnetic field attracted it. You can't have a south magnetic pole without a north one and vice versa. As soon as this flip happened, the vehicle would crash back to Earth hard.
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Finally a way for me to pick up that 7-10 split!  :)\n\nI agree with the basic premise of oscillating assuming a vacuum (since we are assuming no molten core, why not a vacuum as well?).\n\nBut, depending on the direction of the hole in relationship to the rotational axis of the earth, wouldn't the ball start bouncing against the sides of the hole due to the mechanics similar to that of a Foucault Pendulum?
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Two planes take off together from base. 1/8 distance around the globe one fills up the other and goes back to base. \nthe filed up plane by that time is 1/4 distance around the globe. The two planes at the base fill up and take off in the opposite direction of the original flight. at 1/8 the distance one plane fills up the other and returns to base. by this time the first plane is 1/2 the distance around the globe.The now filled up plane flys 1/8 the distance around the globe to meet with the first plane which has an empty tank now.  now we have one plane with an empty tank  and one plane with 3/4 full tank. the planes divide the fuel among them and go to base which is 3/8 the distance away and they land safely.