Falling from Everst?

[Lawle$$]

So how long does falling from Everst take? I heard someone said it only takes 12s but my calculatios shows it takes about 42s! Am I missing something here or that person has made a mistake? I need the real time not the approximate:eyebrow:

Height: 8,848 m (29,029 ft)

 

 

Thanks:-)

[insane_alien]

well, you see you don't fall 8,848m. the slope of the mountain will intervene quite violently sometime before that. also, due to terminal velocity, falling 8.8km would take substantially more than 42 seconds. probably about 2-3 minutes.

[MrSandman]

Wouldn't you reach a maxium velocity, because of wind resistence?

 

Well, it took me 3 seconds to fall from a 40ft cliff, so maybe you could at least get a generalization, also I weigh 130.

 

You can see that you can't get a generalization, because you didn't specify the weight of the object falling.

[timo]

well, you see you don't fall 8,848m. the slope of the mountain will intervene quite violently sometime before that.

Even if it didn't, the Mt Everest does not exactly rise from the sea but from the Himalaya highlands. One could try to guesstimate the height from which Mt Everest rises from the 12 s and the 42 s (which btw. are approximately correct - a free fall over 8 km with a gravitational acceleration 10 m/s² takes 40 s).

[MrSandman]

I think you need to realize terminal velocity. I don't know anything really, but i pretty sure it would take longer than 40ty seconds. Ask cpt_refssmet. Or at least his name is something like that.

[Lawle$$]

Even if it didn't, the Mt Everest does not exactly rise from the sea but from the Himalaya highlands. One could try to guesstimate the height from which Mt Everest rises from the 12 s and the 42 s (which btw. are approximately correct - a free fall over 8 km with a gravitational acceleration 10 m/s² takes 40 s).

 

How would 12s be correct even approximately?

[ydoaPs]

How would 12s be correct even approximately?

 

IIRC, there's not a vertical face of the mountain from peak to base, so you won't fall the whole way.

[timo]

How would 12s be correct even approximately?

I dunno. It's your thread, so: Why shouldn't they?

[YT2095]

it took me 3 seconds to fall from a 40ft cliff

 

I somehow Doubt that!

[insane_alien]

I somehow Doubt that!

 

what he didn't say was it was only 3 seconds from actually arriving at the top of the cliff to him toppling over the edge.

[DrDNA]

I think that in order to answer this question correctly, you would need just a little more information.

For example, how many verticle cliffs are there on Everest, how high is each cliff, how far apart is each cliff, what is your walking speed to the first cliff, for each subsequent space between the cliffs how fast can two men carrying you on a stretcher walk from one cliff to the next before they dump your broken, mangeled, dead body over the cliffs, and how long does it take for the two men to repel down each cliff with the empty stretcher?

You should probably add a few seconds here and there for picking up your dead body and putting it on the stretcher at the bottom of each cliff (total time should depend on how many cliffs)...and you should have it.

[MrSandman]

Well it was darn close when I was counting when I fell. three seconds was from when I jumped till contact point. You should try jumping off of cliffs it can help you understand. I think his Mt. Everest is more like a pole the same height as everest. It is impossible to fall from Everest to sea level.

[Lawle$$]

oh my...you guys are taking the problem way too serious! ok ok I give up:doh:

 

thanks all of you:D

 

I dunno. It's your thread, so: Why shouldn't they?

 

One could try to guesstimate the height from which Mt Everest rises from the 12 s[/color'] and the 42 s (which btw. are approximately correct - a free fall over 8 km with a gravitational acceleration 10 m/s² takes 40 s).

 

perhaps I've misunderstood your post.

[timo]

perhaps I've misunderstood your post.

Most certainly. Quoting me to explain what you meant really doesn't help me understanding what you meant. The 12 s originate from your post #1.

[Lawle$$]

Most certainly. Quoting me to explain what you meant really doesn't help me understanding what you meant. The 12 s originate from your post #1.

 

I know! But I thought you were saying it was correct:-)

[timo]

Ok, slow. I'm not sure if all misconceptions are already cleared up, but just in case they aren't:

What I said was: A free fall (which also means no air resistance) over 8 km does, using a gravitational acceleration of 10 m/s², take 40 s. What I assumed was that you got your 42 s from calculating the time it takes for a free fall over 8,848 m with a gravitational acceleration of 9.81 m/s². Since the numbers are similar, the 42 s seem to be the correct result. What one could imagine now is that the 12 s are the calculation of how long it takes to freely fall over a distance equal to the distance from the very top of mount everest to it's lower end (it doesn't rise from the sea, after all - the exact definition of lower end might still be a problem but visually, you often see approximately where a mountain starts to rise). What I furthermore proposed was that the two numbers (42 s and 12 s) are, assuming they are correct and calculated the way I assumed, sufficient to calculate the height at which Mt. Everest starts to rise from its surrounding highlands.

[Lawle$$]

Ok, slow.

My English is terrible!

 

I'm not sure if all misconceptions are already cleared up, but just in case they aren't:

What I said was: A free fall (which also means no air resistance) over 8 km does, using a gravitational acceleration of 10 m/s², take 40 s. What I assumed was that you got your 42 s from calculating the time it takes for a free fall over 8,848 m with a gravitational acceleration of 9.81 m/s². Since the numbers are similar, the 42 s seem to be the correct result. What one could imagine now is that the 12 s are the calculation of how long it takes to freely fall over a distance equal to the distance from the very top of mount everest to it's lower end (it doesn't rise from the sea, after all - the exact definition of lower end might still be a problem but visually, you often see approximately where a mountain starts to rise). What I furthermore proposed was that the two numbers (42 s and 12 s) are, assuming they are correct and calculated the way I assumed, sufficient to calculate the height at which Mt. Everest starts to rise from its surrounding highlands.

 

Thanks!

[Spyman]

Reasonable base elevations for Everest range from 4,200 m (13,800 ft) on the south side to 5,200 m (17,100 ft) on the Tibetan Plateau, yielding a height above base in the range of 3,650 m (12,000 ft) to 4,650 m (15,300 ft).

http://en.wikipedia.org/wiki/Mount_Everest

 

That would give a "free fall" time in the range of 27 to 31 seconds, without air resistance.

 

 

At 8,750 m (28,700 ft), a small table-sized dome of ice and snow marks the South Summit. From the South Summit, climbers follow the knife-edge southeast ridge along what is known as the "Cornice traverse" where snow clings to intermittent rock. This is the most exposed section of the climb as a misstep to the left would send one 2,400 m (8,000 ft) down the southwest face while to the immediate right is the 3,050 m (10,000 ft) Kangshung face. At the end of this traverse is an imposing 12 m (40 ft) rock wall called the "Hillary Step" at 8,760 m (28,750 ft).

12 seconds corresponds to the approximate height of only ~700 meters, which then clearly must be wrong.

[MrSandman]

Yeah, mind you air resistence is quite powerful. That's why airplanes can fly.

[insane_alien]

no, they don't fly because of air resistance, air resistance is the reason they need to have the engines on all the time.

 

aeroplanes fly due to the bernoulli effect.

[MrSandman]

yes, I know, but if air resistence wasn't present. The bernoulli effect wouldn't work.

[Spyman]

Yeah, mind you air resistence is quite powerful.

That would only make the "twelve seconds" answer even more wrong, wouldn't it ?

[insane_alien]

yes, I know, but if air resistence wasn't present. The bernoulli effect wouldn't work

 

Explain? i see no resistance term in the bernoulli equations. well, they are in the modified equations for flow in a pipe but that doesn't quite apply for planes. also, if there was no resistane the effect would be much greater.

 

the pressure drop is a result of the velocity of flow. the resistance would lessen this flow and reduce the pressure drop.

[Bornsatin]

I would see your reasoning, but whatever the solution maybe, 12 seconds is hardly even rational in this case. Was it one of those "Interesting Facts" that are meant to mislead and impress weak-minded teenagers? THAT would be my conclusion.