By the conservation of energy law, we have TE = PE + KE + W; where total energy is the sum of potential and kinetic energies, and of work done on the system. In the beginning, at height H above ground level, TE = PE only because no work has been done and there is no velocity of fall for KE. But TE in the beginning has to = TE upon impact at ground level because energy is neither created or destroyed. That means TE = constant no matter where the system is in the fall.
The biologist and the egg are that system. Relative to an ouside observer, and as the system falls, TE = PE + KE + W; where PE = Mgh (and h, height above ground level, is diminishing really, really fast), KE = 1/2 MV^2 (and V is increasing really, really fast), and W = MgS = Mg(H - h); where H - h = S the distance the he and the egg have fallen from the start at H above ground.
What happens here, as he and precious fall, is that gravity does work on them both and that work changes their combined potential energy into kinetic energy. This is why as more work is done (i.e., the distance fallen S gets bigger), potential energy gets smaller (i.e., h lessens) and kinetic energy increases as V gets larger.
In the instant before impact at ground level, we have TE = KE because h is essentially 0 and all the work is done because there is no more distance to fall. That is, there is no more work to convert PE into KE. Thus, the impact velocity is simply TE(impact) = KE = 1/2 MV^2 = MgH = PE = TE(start). And we see the familiar V = sqrt(2gH).
Now, our intrepid physical biologist (this is a physics question) tosses the egg of mass m upward with a velocity v upward just before impact. In this case, just as the egg leaves his hand, it has V - v = u velocity relative to the outside observer. As velocities are vectors, they are additive; so ke = 1/2 mu^2 upon release from the hand.
Now, and this is a big NOW, which is bigger V or v? If V > v, the V - v = u > 0 and the egg is still falling downward even though at a slower velocity. If V < v, then the egg, relative to the outside, is traveling upward, back up the shaft at u < 0.
This ke will convert to pe = mgh' as it works against gravity w = mgh'. So, for the egg, te = ke = 1/2 mu^2 = mgh' = pe = te; where te in the total energy of the egg alone, ke is the kinetic energy of the egg leaving the hand and when it returns again after its pe is changed back to kinetric energy by falling back to h' = 0. In other words, lacking air drag, the velodity of the egg u going up will be the same velocity u upon impact.
So V - v = u is critical to saving that egg. If u is too large, the kinetic energy of the falling egg will be too great and smash the egg. So let's look at V = sqrt(2gH) ~ sqrt(20*20) = 20 m/sec, which is about 45 mph. So if our intrepid biophysicist is savvy, he'll launch the egg about 45 mph straight up. In which case te = ke = 1/2 mu^2 = 0 = mgh' = pe = te because the resulting velocity vector is V - v = V - V = 0 = u.
As he and the egg are descending at V = 45 mph, inside our biophysicist has to see the egg leaving his hand at about 90 mph upward from the floor of the elevator. This results because 90 mph up relative to the elevator is only 45 mph upward relative to the outside. This is a major league pitch, but scared as hell, he can do it because of all that adrenaline.
I am guessing, but eggs sometimes shatter when falling off a countertop; and that means an impact velocity of w = sqrt(20*1) ~ 4 mps or 9 mph. So V - v = 45 - v = 9 and 36 mph is the minimum upward velcity relative to the outside for the egg to survive. Thus, it might survive if he tosses it upward 81 mph or more relative to the elevator floor. But that's still a hefty toss for our PhD genius.
PS: While we're doing reality checks, consider what will happen to the roof of the elevator when the floor of the elevator comes to a screeching halt from 45 mph...same thing that happens to the hood of your car when running into an immobile wall at 45 mph...smoooosh. So, in the end, the egg gets smashed by the roof as it collapses to floor level of the elevator. Ah, well.....
2007-09-24 11:43:13
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answer #1
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answered by oldprof 7
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relies upon on how severe the elevator began. in spite of the top, you and the elevator fall on an identical velocity (as Galileo proved), so in case you need to fall from that top (without the elevator in touch) and be ok, then confident. yet while it rather is lots over 10 ft up, then no, of course no longer. those actual impossibilities are what make the cartoons humorous. Wiley Coyote would not fall while he's administered off the sting of the cliff, case in point, until eventually he seems down.
2016-10-05 07:17:48
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answer #2
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answered by mccleery 4
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I think his best bet would be to fall with the egg and take away the force from the egg as he goes down:
http://s223.photobucket.com/albums/dd78/floodtl/?action=view¤t=saveegg.jpg
Well, not exactly like that but you get the picture. He falls and takes the velocity off the egg slowly. He would want his legs to be under him though, so maybe falling backwards would be better.
How's that throwing the egg? He's deccelerating it slowly.
2007-09-24 10:40:45
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answer #3
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answered by Anonymous
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Sure, this way it will brake on the ceiling of the elevator!
Covering the egg would be the best way to attempt to save it, but I doubt it would work.
Then again, even if it is broken, can't he use the egg's DNA?
2007-09-24 10:24:51
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answer #4
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answered by Yahoo! 5
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