What is the information we need to calculate the falling location of a object from a flying aeroplane?
397135
The weight of the object is irrelevant unless one is taking into account air resistance. This is because the net force on the object is the force of gravity, so we have ma = -mg, or a = -g, which is all we need to know for the vertical kinematics equations. The only thing we need to know for the horizontal equation is the initial horizontal speed of the package.\nIf, on the other hand, one was concerned with air resistance, one would also need to know approximate air density and cross-sectional area of the object, as well as be able to numerically solve a nonlinear ordinary differential equation (the frictional force is proportional to some function of velocity) so I doubt this is what you wanted.
460474
HI MAWBY --\n\nI ANSWERED THIS YESTERDAY IN SOME DETAIL.\n\nPLEASE GO TO THE SITE AND QUESTION BELOW.\n\nITS IN THE AILERONS, TABS, AND ELEVATORS.
333602
If you mean by 'above' you measure along the ramp, then the vertical drop is 40sin(30) which is 20 meters.\n\nSo, you just need to figure out how fast an object will be traveling after falling 20 meters (I am leaving off friction here).
1152034
Well, if you know the thrust of the engines, and can measure the acceleration of the aircraft, you can calculate the mass of the plane.
1144696
gravity and air resistance.
459122
the side length is 2\nso the altitude should be 3^(1/2)\ngood luck
614422
Depends what information you already have. The simplest one I can think of is:\n\nVelocity = Acceleration x Time\n\n-Therefore-\n\nAcceleration = Velocity / Time\n\nBut you might have other variables involved here.
181081
That is a very good question.  The weight of the fly is inside the airplane whether it lands or not.  When I asked a similar question years ago, I was told that the weight of the fly is transfered to the surface under it while it is flying.  I then asked: "If that is true, how come when a huge airplane flys over my head I'm not crushed?"  They answered because the weight is dispersed over a large area of the earth.  I don't understand any of that physics stuff.  I will look forward to checking back to see if some egghead will explain it for us.  Good luck.
824568
We can use trigonometry to determine the vertical and horizontal components of the stone's velocity.  This will let us determine how far it goes.\n\nImagine a right triangle with a hypotenuse of 14 units and one angle of 26 degrees.  The side opposite the 26 degree angle will be as long as the stone's vertical velocity.  The side adjacent to the 26 degree angle will be as long as the stone's horizontal velocity.\n\nSince:\n\nsinA = opposite / hypotenuse... \nsin26 = vertical velocity / 14...\nvertical velocity = 6.14m/s\n\ncosA = adjacent / hypotenuse;\ncos26 = horizontal velocity / 14...\nhorizontal velocity = 12.56m/s\n\nIf the object had an initial vertical velocity of 6.14m/s, we know how long it stayed in the air because:\nacceleration = change in velocity / time\nWhich is the same as saying:\ntime = change in velocity / acceleration\nThis applies in any situation where the acceleration is constant.\nSo, if we want to know how long it took for the stone to reach 0m/s vertical velocity (the point at which it stops rising and starts falling), the change in velocity will be 6.14m/s.  If we just divide this number by acceleration due to Earth's gravity (9.81m/s^2), we get 0.63 seconds.  The stone traveled upward for 0.63 seconds.  This means that it must have also traveled downward for 0.63 seconds before hitting the ground, for a total flight time of 1.26 seconds.\n\nNow, we have a simple velocity equation.  Since we know that:\ndistance = velocity * time \nSince we're only worried about how far the stone traveled horizontally, we use the stone's horizontal velocity: 12.56m/s.  Time will be the total time that the stone was moving: 1.26sec.\n\nSo...\ndistance = 12.56 * 1.26 = 15.83 meters.\n\nHope that helped!
204311
It's around 23.6 mph (penny falling flat) to 73.3 mph (penny falling on its side).\n\nVterm =  sqrt ((2 * m * g) / (c * p * A))\n \nwhere:\nm=mass of the object \ng=acceleration from gravity \nc=coefficient of drag \np=density of air \nA=Area of the object \n\nIt's pretty straightforward.  The full explanation is at the source.
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!
449788
Doesn't matter whether it's dropped from a helicopter, skyscraper, or the goodyear blimp.  Terminal velocity is the velocity at which the air resistance of a falling object is equal to the acceleration due to to gravity, and the object can no longer can gain speed.  It's not a constant, as you get closer to the ground air pressure will increase, so the air's resistance will increase, and the teminal velocity will drop. The object will actually slow down.
846846
Very easy.  If you know which direction the object is going, then you know the direction of your velocity vector - the angle (let us say "c") it makes with x axis.  Draw xy coordinates on your paper to represent a plane that contains your velocity vector.  You can draw a line parallel to x axis from the tip of the velocity vector.  That will give you the vertical velocity.\n\nIf the angle is "c", for example, your horizontal velocity is M cos c and your vertical velocity is M sin c, where "M" is the length of your velocity vector, that is it's numerical value.\n\nYou can also
1118371
double_nubbins has the right approach.  I will add a couple of hints: how long will it take the car to fall 19ft (remember the acceleration equation s=.5*g*t^2).  Assuming there is no air drag, the horizontal velocity of the car must be enough to cover the 52m width in that time.\n\nSpeed of the car? The horizontal speed? The falling speed?  The speed in the direction of motion?\n\nHorizontal speed: as calculated above\nFalling speed = g*t, where t is the time calculated above.\nSpeed  in the direction of motion = sqrt(horiz speed ^2 = falling speed ^2)
317623
i believe the formula is\n\n16 foot per second per second\nor\n16ft + 32ft + 48ft + 64ft + ect....\n1 sec, 2 sec, 3 sec, 4 sec, ect....\nadding 16 foot for each additional second\nuntil you are adding the terminal velocity for each second.\nthe terminal velocity is the fastest something can fall.\nthe terminal velocity changes with the shape and mass\nof the falling object.\nexample : a piece of paper vs a sphere of lead\nyour ball would be approx. 96 ft of the way down\nor 4 ft off the ground, provided it wasn't Styrofoam.
680534
Use the equation from the link below.  You will find that \nv/c = 0.14365, or\nv = 0.14365 c\nor\nv = 43,100 km/s\n\nThe object is moving away from the observer at this rate.\n\nBy the way, you probably won't get credit unless you can work out the math steps I left out.
899952
If you throw a ball across the room, it will stop accelerating as it leaves your hand.\n\nThe whole ball (including anything inside it) will carry on travelling according to the force you applied, but its direction will be altered by gravity (which is still pulling it downwards).\n\nThe shape it's flight makes is called a 'parabola', and this same shape happens whenever gravity stops an object from travelling upwards (even a purely vertical thrust will lead to some sideways movement; this is called 'Coriolis effect'), and makes it travel downwards instead.\n\nBecause the outside of the object is always travelling in the same direction as the objects inside it, they don't press on each other (their relative 'G' is zero). Because 'weight' is actually just relative pressure, the objects will seem weightless.\n\nThe American and Soviet 'vomit comets' are large transport aeroplanes with strengthened wing roots and a padded interior.\n\nThe pilot will first initiate a steep dive, and pull up out of this as if he is going to 'loop the loop'. Pulling up from a dive like this will press the passengers against the floor, so they seem heavier than normal; typically 3-4 times their normal weight (3-4G).\n\nAs the plane comes up from the dive, this extra 'G' will decrease, until it reaches zero, and the passengers feel 'weightless'. \n\nAt that point, the aeroplane is travelling upwards under the momentum of its own weight, like a roller coaster. If the pilot kept the engines accelerating, they would drive the plane upwards and press against the passengers, but by cutting the engines, the plane is left to rise like a thrown ball. It travels in a parabolic curve, but the passengers don't (they tend to bounce around in zigzags, perform somersaults for the camera, lose their sense of direction and start puking)\n\nAbout 40 seconds of weightlessness can be simulated in this way, because that's how long it takes for the plane to stop rising, and return to its downward dive speed. If it was left to dive for too long, pulling out of the dive would rip the wings off.\n\n(if that happened, the passengers would experience weightlessness (or very nearly; air resistance would slow the plane, but not them) all the way to the ground, but they wouldn't be able to tell us about it afterwards)