When you jump into the air and land again on ground ,will it b the same point.read details?
613764
I assume you are wondering if the Earth's rotation will come into play here.  \n\nThe answer is yes.  This is the same as asking if a ball thrown in an airplane will land back in the person's hand.  Why is that so?  Everything in the plane is moving at the same speed, even the air molecules inside the cabin.  So, when the ball is thrown up, since it had the plane's speed before it was thrown, it will have that speed while going in the air, and since the air molecules are trucking along with everything else, they won't hinder the ball's motion.  \n\nIf you were to be able to watch the ball be thrown from outside the cabin, you would see the ball follow a curved parabolic path, where everyone in the cabin would see it go straight up and straight down.  You would see that curved path, because of the ball's horizontal motion due to the plane's speed.  Everyone inside wouldn't not see that, because they too are traveling with that same horizontal motion.\n\nSame thing applies here.  You share the same tangential speed as the Earth's surface, and so do all of the air molecules.  If you were able to jump for that long of time, you and everyone else around you would see you go straight up and straight down.  However, if you were being watched from space, you would follow a parabolic curve because of your tangential motion.
1006314
No.  When you departed the aircraft you body would be travelling at the same speed as the aircraft was travelling.    Wind resistance would slow your body down very quickly but you would still land well ahead of the point in which you jumped out of the plane.
747360
Well gosh....  \n\nsince the only linear unit mentioned in the question is 18 ft/s  let's assume that the 60 is going to give us acceleration in ft/sec^2 \n\nSo, the position function for this is the second integral of the acceleration function, right? \n\nso...  the first anti-derivative of 60t is  30t^2   which gives the velocity, so at the end of the boost phase, our rocket is moving 30 * 3 * 3 = 270 ft/s \n\nand the second antiderivative is 10t^3  right?  \n\nSo, the position at time T is 10 * 3 * 3 * 3 = 270 feet.  (odd coincidence, but there you are...) \n\nNow, the rocket begins a SECOND acceleration by gravity of -32 ft/s negative 'cause we're accelerating DOWN now... ....   but this time the acceleration is constant,  so \n\nV = gt   and s = 1/2  a  t^2 so, we're slowing from 270 feet per second to zero at the top...  and  t = v/g = 270 / 32 = 8.44 seconds and the height we coast to is the 270 we had at the end of boost plus   1/2  *  32  * 8.44 * 8.44 = 1140 feet.  \n\nThe parachute is going to open after 14 seconds...   so, 14-8.44 is 5.56 seconds of free-fall (and we'll ignore air resistance again) so, we go from zero to something in 5.56 seconds....   and position is 1140 - ( 1/2  * g  * t^2 ) again...   so, the parachute opens at \n\n1140 - (1/2  * 32  * 5.56 * 5.56 )  = 645 feet. \n\nad the velocity is  gt = -32 * 5.56 = -178 feet per second...  \n\nNow, for the NEXT five seconds, we slow under the effect of the parachute   to  -18 ft/sec -- so,  we slow by 160 feet per second in five seconds, so the acceleration of the parachute is (oddly) 32 feet/second squared (huh, how odd is that? )  \n\nSo, the distance traveled during that phase is Vo t + 1/2 at^2  \n\nSo, we have (-160 * 5) + (1/2*32*5*5) =  -800 + 400 = 400 feet.  \n\nand from 400 feet up, we continue down at 18 ft/s for an additional  22 seconds...  \n\nand the rocket lands at t = 22 + 5 +  14 + 3  = 44 seconds \n\nAnd yes, I dropped a bunch of decimal points.  Your original data was only to two significant digits....
81552
If you disregard air resistance, both land at the same time.\n\nThe Earth's gravitational pull affects everything equally.
128500
The object will follow a parabolic trajectory to the earth, as observed from outside the plane.  On the plane, it will appear to fall straight down in your (moving) frame of reference.  Air resistance will retard this motion, but it's magnitude depends on the drag coefficient of the object dropped and the medium in which it is falling.
84559
Both marbles will hit the ground at the same time, but one will roll along farther.\n\nIn short, the vertical force from gravity is the same for both marbles.  The one hit sideways just has an additional horizontal velocity in addition to its vertical acceleration.
496012
javelin thrown at angles 30 and 60 will land in same spot because as per properties of projection if x is angle of projection then range for angles of projection x and 90-x will be same
526447
I actually had my class conduct an experiment on this very thing.  The problem was: does the percentage of energy lost from a ball bouncing increase, decrease, or stay the same as the height increases.  This is akin to your question, because the height at which it bounces back to depends on how much energy is left after the bounce.  \n\nWe found that the percantage of energy lost stays a constant, which means the the answer to your question is yes.\n\nFor example:\nIf you start with a height of 10 meters and drop a 1.0 kg ball, it will have 98 Joules of potential energy (PE = mgh).  If it loses 10% of its energy on the bounce, it will have 88.2 Joules of kinetic energy, which will translate to a height of 9.0 meters.  \n\nIf you double the original height, you will have 196 Joules of energy.  After losing 10% from the bounce, you are left with 176.4 Joules, which translates to a height of 18 meters after the bounce.
744790
Yes. That is called lucid dreaming, when you realize you are dreaming. Next time, try flying, it's pretty fun. Careful though, you could lose control, mid control and start flying upside down or hovering just above the floor! Watch a movie called "Waking Life" It's all about dreams and its a very, excellent and entertaining movie! One of my faves! :)
1069059
Ok well lets assume that we are not taking pressure or temperature into account, nor friction due to to air speed etc. If you jumped in the hole you would fall, and continue falling at greater and greater speed. Now as you approach the center of the earth the force of gravity would decrease  but  momentum would keep you going right thru the center and  towards the other side.\nNow gravity would start to increase again as you get farther from the center, but it would tend to start slowing you down. until it caused you to stop, right about the other side of the hole in the earth. Think of a bungee jumper, same thing. Now if you were not able to hold on to something you would start falling back down the hole and it would repeat.\n\nOf course this is a VERY simplified view of what would happen. In reality  you have things like the rotation of the earth th consider, which would cause you not to fall straight down the hole, but into the side wall of the hole. But it is a good thought experiment to help understand gravity.
71702
If the only factor involved in this would be gravity, the bullet would come back down at the same speed as it went up. But since there are other factors (friction, air, etc), the bullet would come down with less speed than it went up.\n\nThe same would happen with the place where it was fired, it probably wouldnt land on the same spot, since other factors have an effect on the bullet trayectory.\n\nchech the link from digg and wait for the episode of Mythbusters where they solve this myth ...
335818
well...throw?..if you dropped them at the same time they will hit the earth at the same time
493374
You shoulf face forward (so that you face the direction in which you will be moving with respect to ground when you land), but push yourself backward (to decrease your velocity relative to ground).
212614
the wall stops it and the ground starts it again. if the spinning force is greater than the stopping force it would continue to spin after hitting the wall.
404977
Hahahahahahaha, how amusing the previous anwers are.\n\nThe balloon would continue to rise in the atmosphere until it reached a height where the pressure inside the balloon was equal to the pressure outside the balloon...then it would remain at the same height, moving in the wind, maybe slightly up and down due to thermals...until it burst !
837767
Bingo!!! You got it...air resistance.  If you dropped them in a total vacuum, they would both fall at the same rate.
971235
Depends on your momentum when you do so.\n\nIf you're utterly still and blow out air in space, you will move backwards.\n\nWell, your frozen corpse will...