When you jump into the air and land again on ground ,will it b the same point.read details?
611220
This is about conservation of momentum. Any object on a rotating (and for our experiment atmosphere-less) planet has the same rotational momentum and angular velocity  as the surface. If it jups up, it will increase the radius, and therefore reduce the angular velocity, so it will land further behind.
837767
Bingo!!! You got it...air resistance.  If you dropped them in a total vacuum, they would both fall at the same rate.
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! :)
817165
Very simply:\n\nIf he keeps flying at the same speed, he will be moving at his flight speed plus the speed of the bus.\n\nIf he stops flying, he will fall to the floor (or perhaps land a bit more gracefully than that), and be moving only at the speed of the bus (his flight speed has gone to zero).\n\nIf you jump in a bus, you continue to move forward at the same rate as when you were standing, so you land in the same spot on the floor.  Of course, if the bus changes velocity while you are in midair, your motion will be different from its and you will not be in the same spot when you land.\n\nThe same goes for the rotation of the earth... the circumference of the earth is about 24,000 miles at the equator.  Hence, at the equator, it is moving at about 1467 ft/sec.  If you jumped and stayed in midair for 1 second, the earth would travel 1467', but you are travelling at the same rate and you would land in the same spot you jumped from.
1137769
plane starts gliding...trading potential energy of its height to kinetic energy forward to hopefully find a landing site.  this is why airplanes gain altitude as quickly as possible after take-off, so if there is an engine problem they have time to find a landing place.\n\nIf thrust is lost, gravity and lift become unbalanced with gravity winning because altitude must be lost to maintain airspeed.  drag wins over thrust since there isn't any thrust so altitude must be lost to compensate for drag-thrust imbalance.\n\nsource: i'm a pilot
411708
of course you can,just like you can jump infront of a moving bus...the result would be the same..
669863
I imagine very slightly, For every action has an equal and opposite reaction. so the force of the body hitting the ground would be the same for the ground would hypothetically push back on the body, But for the most part the bones would absorb the impact, thusforth by breaking, so in a word, yes, but probably not how you'd imagine.
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....
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 !
192657
No way to know.  What is the shape of "everything"?  I think it is round and that if you go to one end, you'll end up back at the same point if you keep going the same direction.\n\nGood question.
1013783
Absolutely. Surround yourself with people that care and they are sure to catch you every time. Besides......once you learn from your mistakes its amazing to see how one can "bounce" back into good graces. So even then, hitting the ground only stings for a second.....then you are off and running and soaring higher than ever.
423059
It is the same.  Unless you're flying in one of the experimental planes which leaves the atmosphere.  The atmosphere moves with the planet so the plane is unaffected by the spinning.  Just as we don't feel the spinning.
200232
depends on the kind of fall and the trajectory
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.
243205
Yes it can hit the ground. Its the players job to prevent it from hitting more than once.
1163141
yes .. in vacuum, all items will fall at same rate... it is the rate at which the gravity acceleration attacts everything to the center of teh Earth. But you might have a queation, that if the acceleration  is same then why different things weigh different... \n\nweight =mass X acceleration of gravity (9.8meter/second^2) and mass is the one which is a measure of matter in a body makes the difference... so if you let a 5kg weight and a 5,000kg weight fall in a vaccum, they will fall at the same rate and will reach the ground at the same time.
1108579
NO!!!\n\n\nThey dont fall with the same velocity, \n\nThere is a maximum /terminal velocity an object can attain while moving in a fluid. (in our case - air).\n\nAnother force acts on bullet in a direction opposite to that of g. \n\nThis force = 6 * pi * n * r * v \n\nr - radius of object,\nn - coeff of viscocity\nv - terminal velocity.\n\n\nThus the velocity with which the bullet will return to earth would be much less. (but still large enough )