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In the movie Frozen, the following dialogue takes place:

Anna: "It's a hundred-foot drop."

Kristoff: "It's two hundred."

Anna: "Okay, what if we fall?"

Kristoff: "There's 20 feet of fresh powder down there. It will be like landing on a pillow... Hopefully.

Then they fall all the way to the bottom and survive.

My question is this: would this be actually possible? My instinct tells me no, but I'm too awful at physics to back it.

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    $\begingroup$ This is difficult because it depends on the properties of fresh powder snow. We need a snow expert. As far as the basic non-snow-dependent physics is concerned you can assume they are travelling at about 50 meters per second when they hit the snow. $\endgroup$
    – hft
    Commented Feb 25, 2015 at 23:02
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    $\begingroup$ Possible, yes. People have survived falling over 10,000 ft without a working parachute, e.g. en.wikipedia.org/wiki/Nicholas_Alkemade and en.wikipedia.org/wiki/Ivan_Chisov Certain? I doubt it. You'd need Goldilocks snow: too hard and you go splat, too soft and you just zip through and go splat on the ground underneath. $\endgroup$
    – jamesqf
    Commented Feb 26, 2015 at 3:09
  • $\begingroup$ As far as I know a pile of snow does not have a uniform density and structure from top to bottom. Typically snow consists of different layers that can differ very much from each other in terms of quality. I remember someone stating in a documentary you are likely to survive a 40 meter (~130 foot) drop if the layering is favorable. For more on snow stability – in the context of avalanches – see Jim Frankenfield's paper. $\endgroup$
    – Søren D. Ptæus
    Commented Feb 26, 2015 at 14:36
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    $\begingroup$ They survived because it was magical snow created by Elsa's magic winter. $\endgroup$
    – Michael
    Commented Feb 26, 2015 at 16:25
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    $\begingroup$ Here's a field experiment a Russian completed. Plus his legs were on fire, just because. youtube.com/watch?v=uRHyVT8F834 . A (translated) report said "...But things did not go according to plan - Alexander could not conveniently grouped and fell on his side, received serious injuries. At the scene promptly arrived MOE, police and ambulance. The preliminary diagnosis - bruised lung, as well as damage to internal organs - told paramedic ambulance. - Requiring hospitalization $\endgroup$
    – Fractional
    Commented Feb 26, 2015 at 22:27

5 Answers 5

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As a very rude guess, fresh snow (see page vi) can have a density of $0.3 \ \mathrm{g/cm^3}$ and be compressed all the way to about the density of ice, $0.9\ \mathrm{ g/cm^3}$.

Under perfect conditions you could see a 13 feet uniform deceleration when landing in 20 feet of snow, or about 4 meters.

enter image description here

Going from $30\ \mathrm{m/s}$ to $0\ \mathrm{m/s}$ (as @Sean suggested in comments), you'd have $(\frac{4\ \mathrm m}{12.5\ \mathrm{m/s}})$ = 0.32 seconds to decelerate.

The acceleration is $\frac{30\ \mathrm{m/s}}{0.32\ \mathrm{s}}$ = $93.75\ \mathrm{m/s^2}$. That's about:

9.5G's of acceleration

Wikipedia lists 25g's as the point where serious injury/death can occur, and 215g's as the maximum a human has ever survived.

So it seems plausible.

But it should be noted that since the snow at the bottom is under a lot of pressure from the weight of the snow above, it's likely the density would not be $0.3\ \mathrm{g/cm^3}$ throughout. It would help that the force lasts only a fraction of a second.

Edit as pointed out in the comments, the force that the snow will exert could vary with its density. So initially, the force would be rather weak, and as you approach $0.9\ \mathrm{\frac{g}{cm^3}}$ that force would increase, probably exponentially. So the above answer is really a "best case scenario" when it comes to snow compressibility

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    $\begingroup$ They fell 200 feet - the 20 feet refers to the depth of the snow. $\endgroup$
    – Dave Coffman
    Commented Feb 26, 2015 at 0:12
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    $\begingroup$ @pentane Average speed for uniform deceleration to 0 is always half the initial velocity. $\endgroup$
    – Señor O
    Commented Feb 26, 2015 at 0:30
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    $\begingroup$ I'd like to point out that @hft 's estimate of 50m/s is incorrect. If Anna and Kristoff are correct in their estimation that the cliff is 200 feet high, that's a height of 60 meters. If you solve for final velocity by using $\Delta y=1/2gt^2$ and then $v_f=gt$ (since initial velocity is approximately 0, you get a final velocity of approximately 35 m/s. Furthermore, that calculation assumes no air resistance, so they'll be traveling at some final speed less than that when they hit the ground, making survival even more plausible. $\endgroup$
    – Sean
    Commented Feb 26, 2015 at 14:53
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    $\begingroup$ The paper you link is not data for fresh snow but rather for compacted snow. This USDA page suggests that the density of fresh snow is more like 0.05 - 0.2 $g/cm^3$. $\endgroup$
    – Nate Eldredge
    Commented Feb 26, 2015 at 22:04
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    $\begingroup$ Why do you assume that deceleration is going to be in any way uniform? The snow doesn't "know" that it's going to take 4m to decelerate, even if the snow is uniform density. They could decelerate to zero in the top 10 cm, or they could be still falling at nearly full speed when they've penetrated 4 m, then come to a sudden crushing halt. I wouldn't say your answer makes it in any way plausible; it just places a lower limit on the maximum acceleration, and showing that it's not impossible based on the distance alone. It doesn't tell us anything about how high the acceleration will be above 9.5G $\endgroup$
    – Lodewijk
    Commented Feb 26, 2015 at 23:04
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@Señor O gives a very good answer, but he assumes an ideal deceleration. Based on a viewing of the scene, Anna sinks a little under a meter, while Kristoff doesn't sink more than half a meter.

Since they fell about 200 feet (about 60 m), my initial estimate for their impact velocity is (assuming no air resistance):

$v = \sqrt{2gh} = \sqrt{2*60*9.8} \approx 35 \ \mathrm{m/s}$

However, using a handy chart found in the resource below, when we factor in air resistance, Anna and Kristoff's impact velocity is actually around $33 \ \mathrm{m/s}$

In Kristoff's case,

$v^2 = v_o^2 + 2a\Delta x$

$1100 = 2(0.5)a$

$1100\ \mathrm{ m/s^2} = a$

which is about $110g$. Possibly fatal, especially considering that the way he lands would cause severe stress on the spinal cord.

In Anna's case,

$1100 = 2(1)a$

$550\ \mathrm{m/s^2 }= a$

which is about $55g$. Probably survivable, (some car crashes experience higher gs), but would likely injure her. She does land feet-first (probably the optimal way to land in this case), which would prevent some injury. In short, the duo might survive, but they would not be able to just get up and continue on their merry way.

This FAA paper is my primary source for my calculations.

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  • $\begingroup$ You might need to add air drag to the calculation, as it's not obvious that this fall speed is below terminal velocity. $\endgroup$
    – gerrit
    Commented Feb 26, 2015 at 16:23
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    $\begingroup$ @gerrit Will do - my first inclination was that, given how rough these estimates were, it wouldn't matter much. $\endgroup$
    – Dave Coffman
    Commented Feb 26, 2015 at 21:58
  • $\begingroup$ how do you factor in air resistance? $\endgroup$
    – aloisdg
    Commented Sep 5, 2017 at 21:51
  • $\begingroup$ @aloisdg I used a chart in the linked FAA paper. It contains a plot giving the velocity at impact as a function of fall height for humans. Since it's based on real-world data, air resistance is taken into account. It's not a particularly elegant method, but it works well enough considering how approximate these numbers are. $\endgroup$
    – Dave Coffman
    Commented Sep 6, 2017 at 18:03
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    $\begingroup$ @aloisdg As far as I can tell, there is no given formula. I merely read the graph to come up with my numbers. $\endgroup$
    – Dave Coffman
    Commented Sep 7, 2017 at 20:14
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This is another chance to use one of my favorite approximations ever! I first offered it as an answer to a question about how deep a platform diver will go into the water. Now is the chance to use it again!

Issac Newton developed an expression for the ballistic impact depth of a body into a material. The original idea was expressed for materials of approximately equal densities when the ballistic body is moving fast enough for the target material to behave as a fluid (think cannon ball into dirt, meteorite into lunar regolith, etc). For a human body into snow, we can assume it will behave in a granular enough fashion.

The human body has a density of roughly $985 \ \mathrm{kg/m^3}$. Using the two limits for the density of snow provided in another answer to this question, snow has a density between $300$ and $900\ \mathrm{ kg/m^3}$. Let's assume the characters are 5 feet tall (you can easily change the number used, it's not a complicated formula). This gives us two limiting expressions:

$$d = 5 \times 985/300 = 16.4~\text{feet}$$

and

$$d = 5 \times 985/900 = 5.5~\text{feet}$$

So it really would depend on the actual density of the snow, but if you assume that it starts out around $300 \ \mathrm{kg/m^3}$ and can reach a maximum of $900\ \mathrm{ kg/m^3}$, we can assume that the final depth will be close to the same as assuming the average value as $600\ \mathrm{ kg/m^3}$ which would give:

$$d = 5 \times 985/600 = 8~\text{feet}$$

That will give you a pretty good idea of the penetration depths over that range of densities. These numbers are all pretty close to what is given assuming the ideal deceleration given by this answer.

If you want to do a much more complicated analysis of penetration depth, check out the other, more detailed answer to the platform diver question. There it is actually shown that the penetration depth approaches this Newtonian approximation pretty well! It is also interesting to note that the penetration depth does not depend on the impact velocity/original height. Assuming one is going "fast enough" for the material to behave like a fluid, the expression seems to hold.

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    $\begingroup$ "the girls are 5 feet tall" - Anna is female, Kristoff is male, and listed as "tall": disney.wikia.com/wiki/Kristoff $\endgroup$
    – mskfisher
    Commented Feb 26, 2015 at 12:01
  • $\begingroup$ @mskfisher Yeah... Like I said, never saw it and my ignorance is showing... $\endgroup$
    – tpg2114
    Commented Feb 26, 2015 at 17:02
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Nice theoretical answers (I can certainly appreciate them, I'm a mathematician). But why delve into theory when experiment is available? In this video you can see a skier jump from more than 200 feet and get head first into the snow, without a helmet.

The video starts with the aftermath, if you want to see the jump right away fast forward to about 1 minute into it.

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  • $\begingroup$ Spoiler: the linked clip title indicates this was a planned world-record attempt. He goes in head-first and is buried far enough that only his skis are showing. The voiceover from the paramedic (who was standing by) is that he split his lip but is otherwise fine. The clip ends with the skier slaloming off down the rest of the hill. $\endgroup$
    – brichins
    Commented Sep 5, 2017 at 21:59
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About 50 years ago in Reader's Digest there was an article about a Soviet airplane pilot who bailed out at high altitude. He fell into a snow-filled ravine and survived. If the angle of the snow is high enough it is no big deal. At Squaw Valley I have seen skiers do drops that might have been 100 feet. If the landing is steep enough it is OK. It is "flat landings" that will get you.

Rock climber Lynn Hill fell 100 feet onto a dirt slope. She not only survived, she recovered completely.

Stunt men do quite high jumps onto airbags. 100 feet onto 20 feet of snow seems possible, but I wouldn't try it if I had any alternative.

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