An aeroplane flies approximately 2,3 kilometres above you.An object falls off, how far will it be from you?
400216
It depends on windspeed and direction, and the planes direction in relavence to that speed and direction. It depends on the speed of the plane. It depends on the density of the object, because of wind resistance. EX. feather would be close, lead ball would be further. Your question is like me saying, "I am in a car, how long will it take me to get to McDonalds?" Not enough info
341507
You mean find "g" as if you didn't know the constant? Okay.\n\nPretty simply really. Have the height measured that you're going to drop it from to where it's going to land (this will be your "h"). Make it a high one (one meter is enough). Measure the mass of the object (that's the "w"). Then when you do the experiment, time it. Start it at the instant you let it drop and stop it once it touches where ever it lands. That's enough information to find "g." Just solve for "g" then that would be the answer.\n\nActually, for an experiment, you should solve for "g" as a formula before you start doing anything else. So the formula would be like this:\n\ng = (2h)/[(t^2)(w)]\n\n(t^2 is "t squared")\n\nSo just have that formula then do the experiment: measure height of the drop, the mass of the object, then time it when you drop it (you need good reflexes when you drop it, as it falls fast).\n\nPS Penguins!!!
63833
50 feet or about 2-3 seconds
677923
Drag is the force of the air which interferes with the speed of the plane
824167
The radius is actually 2/3 of the altitude.  Try that
1161605
If the aircraft is pulling a load factor of 2.5 then the wing must produce 2.5*5000 kg or 12500 kg.
613561
Image in a plane mirror will be at the back sidekof the mirror at a distace equal to the distance in which the object is infront of the mirror.\n\nHence the answer is 15 mm.
61716
The air inside the vehicle is moving the same speed as the vehicle. To the fly it's still air, unless a window is open, then the fly may not be able to fly normally.
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!
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
768444
Plane upwards force > then Earth gravitational pull + mass of the plane+how fare it want to be escape.
908785
very close to 2 meters.
217674
3 feet or 36 inches or 0.9144 meters
1124964
To answer this question, you must tell us how fast you are flying.  The distance from pole to pole is approximately 20,025 km.  So the flight time would be:\n\nt = 20,025km / speed\n\nFor example, at the cruising speed of a Boeing 747-400ER of about 1050 km/h, you would require:\n\nt = 20,025km / 1050km/h ~= 19 hours
831721
Are you sure it didn't just fall off somewhere...? Or that it might be a factory defect...? \n\n22 NOV 06, 0354 hrs, GMT.
899525
Using the example....from the problem you can tell that the tail wind helps the plane move faster.  Let's say the speed of the plane in still air is x, so the speed of the plane with the tail wind is equal to x+25.  The distance the plane travels is equal to (x+25)5, which is the speed of the plane times the number of hours the plane flew.  Now for the return flight, we know the wind is now working against the plane, so it reduces the plane's speed.  Therefore the speed on the return flight is x-25.  The distance traveled can be calculated by (x-25)6.  Since the flight to the destination and the return flight must be the equal, we can set the  distances equal to each other.\n\n(x+25)5 = (x-25)6\n5x + 125 = 6x - 150\n\nFrom there you can solve the problem.
1135263
The height of the trajectory is 1 m (3.05-2.05)\n\nThe horizontal distance is 6.02 m\n\nNow you have ucos(25)t = 6.02\n\nand usin(25)t-0.5*9.8*t^2 = 1\n\nSubstitute t = 6.02/ucos(25) in the 2nd eqn and solv.