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If I jump from an airplane straight positioned upright into the ocean, why is it the same as jumping straight on the ground?

Water is a liquid as opposed to the ground, so I would expect that by plunging straight in the water, I would enter it aerodynamically and then be slowed in the water.

user2428118
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Conrad C
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    Related: http://physics.stackexchange.com/q/106826/29216 – BMS Apr 04 '14 at 16:27
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    It won't be exactly as same as a land crash, but fatal nonetheless; and I'm surprised no one has used the word impulse in their answers yet! – Renae Lider Apr 04 '14 at 17:42
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    Including the [tag:experimental-physics] tag on this question seems rather morbid. – jpmc26 Apr 04 '14 at 19:14
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    Being unconscious and severely wounded on land is bad. Being unconscious and severely wounded on the open ocean really sucks though. – Christian Apr 04 '14 at 21:14
  • @Christian look at my answer for a detailed explanation that neither being severely wounded, nor being unconscious under water is a real problem in this context. The point is: it's just far worse. :) – Volker Siegel Apr 04 '14 at 21:32
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    @VolkerSiegel Well, I'm no doctor but I believe shattering your bones can disperse a considerable amount of energy. I mean there are insane falls that people have survived: https://en.wikipedia.org/wiki/Vesna_Vulovi%C4%87 http://www.veteranstoday.com/2011/05/09/survival-f-8-crusader-pilot-falls-15000-with-failed-chute-and-lives/ – Christian Apr 04 '14 at 22:05
  • Take look at section "2009 report" in that page - looks like it was more like "a few hundred meters". Makes more sense to me. – Volker Siegel Apr 04 '14 at 22:15
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    There are at least a handful of cases on record of people falling from airplanes (at relatively low altitude) and surviving, some after hitting water, and some hitting more or less solid ground. And 33 people have survived the jump from the Golden Gate Bridge, a height of about 245 feet. – Hot Licks Apr 06 '14 at 02:58
  • @HotLicks You need about 450m to achieve terminal velocity. So most planes should do but the Golden Gate Bridge not so much. – Christian Apr 10 '14 at 22:30
  • @VolkerSiegel Well, see the "See also" section of the Wikipedia article and the other link I had posted. – Christian Apr 10 '14 at 22:35
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    I had a look at Cliff Judkins's story from the second link - if I read it right, his parachute did not open, but eh was still connected to the parachutes lines, the main parachute that was still folded mostly, and the pilot parachute, that is used to pull out the main parachute. In terms of physics this is really far from free falling. – Volker Siegel Apr 10 '14 at 23:05
  • The thing to do, when you're falling through the air without a parachute, is to spread-eagle as much as possible. This slows the body to about 60 mph, vs 2-3 times that speed when falling feet-first. Then, of course, at the last possible moment before entering the water, you assume a vertical position. Simple! – Hot Licks Aug 13 '14 at 17:57
  • (And if you're not positioned directly over a body of water, but one is within view, a falling human (without "bird suit") can achieve about a 1:1 glide ratio, so you should aim for the water.) – Hot Licks Aug 13 '14 at 20:29
  • why is it the same as jumping straight on the ground What does it mean for "it" to be "the same?" –  Nov 09 '17 at 00:42
  • Aerodynamic? Hydrodynamic? – Sovereign Inquiry Dec 16 '20 at 15:23

6 Answers6

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When you would enter the water, you need to "get the water out of the way". Say you need to get 50 liters of water out of the way. In a very short time you need to move this water by a few centimeters. That means the water needs to be accelerated in this short time first, and accelerating 50 kg of matter with your own body in this very short time will deform your body, no matter whether the matter is solid, liquid, or gas.

The interesting part is, it does not matter how you enter the water—it is not really relevant (regarding being fatal) in which position you enter the water at a high velocity. And you will be slowing your speed in the water, but too quickly for your body to keep up with the forces from different parts of your body being decelerated at different times.

Basically I'm making a very rough estimate whether it would kill, only taking into account one factor, that the water needs to be moved away. And conclude it will still kill, so I do not even try to find all the other ways it would kill.

Update - revised:

One of the effects left out for the estimate is the surface tension.
It seems to not cause a relevant part of the forces - the contribution exists, but is negligibly small. That is depending on the size of the object that is entering the water - for a small object, it would be different.

(see answers of How much of the forces when entering water is related to surface tension?)

Community
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Volker Siegel
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  • You forgot about surface tension. That makes a huge difference too. The surface tension makes the surface of water act like a solid on very short time intervals. If you have something hit the water ahead of you to break the surface, it greatly eases the impact. But from an airplane, you'd probably still be dead – Jim Apr 04 '14 at 14:52
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    Yes, indeed, forgot about that. But I think that's not too bad, as I forgot or ignored just about anything except moving mass out of the way. Without even caring about whether it's fluid, actually. The interesting point I'm trying to make is: Just moving away 50kg of resting mass in that "short time" of some milliseconds will kill that person, no matter what kind of matter [sic]. – Volker Siegel Apr 04 '14 at 15:01
  • Regarding the surface tension - I assume gas has no surface tension; So let's assume one is jumping into compressed gas of the same density like water (ignoring that the gas may decompress etc.). If the above is right, it should still be deadly. Not really sure, so may be surface tension is important. – Volker Siegel Apr 04 '14 at 15:06
  • Hmmm, I can't really visualize that. Are you jumping into it from a vacuum? Or from an uncompressed gas? I'd also point out that the space shuttle doesn't shatter upon impact with the atmosphere, although that gas it much less dense. And it can bounce off the atmosphere it the angle is wrong. But nonetheless, we can all agree, it's deadly regardless. I only brought up surface tension because it's the reason small jumps (like belly flops) hurt so much – Jim Apr 04 '14 at 15:13
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    Mentioning that, I'd like to point out that it does matter how you enter the water. High divers need to train to dive properly so they don't hurt themselves. They must learn how to break the surface and enter the water. If you were to belly flop off of a high-diving tower (or from those really high places you see in the shows), you'd probably splat on the surface (I believe "splat" is the technical term). Clearly, how you enter makes a difference as much as how much water you are accelerating – Jim Apr 04 '14 at 15:26
  • @Jim you got me thinking about surface tension - just asked a question about it: http://physics.stackexchange.com/questions/106826/how-much-of-the-forces-when-entering-water-is-related-to-surface-tension – Volker Siegel Apr 04 '14 at 15:31
  • Does the fact that water will not compress have any bearing on the effect? When the space shuttle re-enters, the atmosphere compresses in front of it releaseing a lot of energy as heat. – TecBrat Apr 04 '14 at 15:49
  • Yes, the incompressibility is one of the many things I do not take into account, concluding "it will kill anyway because of this basic effect", Maybe I can make that approach more clear. – Volker Siegel Apr 04 '14 at 16:00
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    I it does matter if it is liquid or gas you enter. Or to be more precise: the density matters. The higher the density of the material you jump in, the faster the deceleration and the stronger the force acting on you. – André Apr 08 '14 at 14:49
  • Are you sure that however you land, it's always as dangerous? It seems like landing feet first and trying to move as little water as possible is a lot safer than belly flopping. – JobHunter69 Jun 21 '17 at 19:52
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    Interesting point! So in this case, the body is very "streamlined". That means it can move easier in fluids. That is, in both gases and liquids. Imagine the person is falling feet first for some distance before touching water (after falling in non-streamlined posture before). The reduced air drag would increase the speed at impact quite a lot, as it is in the region of most dense air. And then [...] – Volker Siegel Jun 21 '17 at 23:48
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    [...] after impact, the streamlined posture makes the person decelerate at a slower rate - which is helpful to survive. But it may be too late: The actual danger may be caused not from rapid deceleration once immersed in water, but from - much more rapid - deceleration when piercing the water surface, experiencing the effect of surface tension. But then, feet first is the best he can do to handle the surface tension. (I did not do the calculation to compare the effects.) – Volker Siegel Jun 21 '17 at 23:56
  • Note: looking at answers to How much of the forces when entering water is related to surface tension? it seems surface tension is not relevant here. – Volker Siegel Jun 22 '17 at 00:01
  • dv.
    1. did not describe terminal velocity
    2. did not mention impulse
    3. did not provide figures supporting "it does not matter how you enter the water"
     – JMLCarter Oct 18 '17 at 09:29
  • The explanation is good but it lacks verifications. I wish this answer would include some numbers. How do we really know someone can't stand that displacement with the minimum speed achievable before impact? – Simon Aug 25 '21 at 00:05
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Let's look at this another way: you're just moving from one fluid to another. Sounds harmless, right? By specification of the problem, we're at terminal velocity when we hit the water. The force of drag (in both mediums) is roughly:

$$ F_D\, =\, \tfrac12\, \rho\, v^2\, C_D\, A = \rho \left( \frac{1}{2} v^2 C_D A \right) $$

You can imagine that everything except for the density term is the same as you initially transition from the air medium to water. This isn't perfectly accurate, because these are very different Reynolds numbers, but it's good enough for here.

That means that the force (and correspondingly, acceleration) will simply change by the same factor that the density changes by. Also, we know the original acceleration due to drag was 1g, in order to perfectly counteract gravity, which is the definition of terminal velocity. That leads to a simple estimation of the acceleration upon hitting the water. I'll assume we're at sea level.

$$ \frac{a_2}{a_1} = \frac{ a_2 }{1 g}= \frac{ \rho_{H20} } { \rho_{Air} } = \frac{1000}{1.3} \\ a_2 \approx 770 g $$

The maximum acceleration a person can tolerate depends on the duration of the acceleration, but there is an upper limit that you will not tolerate (without death) for any amount of time. You can see from literature on this subject, NASA's graphs don't even bother going above 100g.

Note that a graceful diver's entry will not help you - that's because an aerodynamic position also increases the velocity at which you hit.

Alan Rominger
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    Even a "graceful diuver" entering vertically, toes ahead, will suffer from the following: The toes start accelerating with $770g$ about $0.01s$ before the head enters; per $\frac12at^2$ they have moved up (relatively) by $\approx 37cm$ then and move with $\approx 75m/s$ towards the head. That doesn't sound healthy. – Hagen von Eitzen Apr 05 '14 at 18:18
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    @HagenvonEitzen And answerer. Gracefully diving can help. You change aerodynamical profile shortly before hitting the water. This means that you have not yet accelerated from 50m/s to 150m/s when you hit the water, which means that the drag is less than 1g in the air and thus less than 770g in the water. Since drag is proportional to the square of the velocity, if you do this perfectly you would "only" experience 86g, Here switching perfectly means switching from a position with 50m/s terminal velocity to a position of 150m/s in the instance you hit the water. – Taemyr Jun 11 '15 at 08:21
  • @Taemyr The only missing information from this thread would be the amount of time we experience this 86g and the effect on our body, to be able to conclude if a "perfect landing" is possible to survive in water. – Simon Aug 25 '21 at 00:20
  • To give some kind of reference, according to John Stapp's work: "Stapp demonstrated that a human can withstand at least 46.2 g (in the forward position, with adequate harnessing). This is the highest known acceleration voluntarily encountered by a human, set on December 10, 1954." – Simon Aug 25 '21 at 00:38
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Consider jumping into a swimming pool. Do a barrel-roll (sorry I mean cannon ball, that just kind of slipped out). It's fun, you enter the water nicely and make a huge splash, probably soaking your sister in the process (that'll learn her). Now do a belly flop. Not as fun. You displace exactly the same amount of water in the same time, but this time there is a lot more pain and you come away with red skin and maybe some bruising. The difference? You cover more area in a belly-flop than a cannon ball.

At extreme velocities, accelerating your body's mass of water will kill you anyway. However, what actually kills you is hitting the surface. Dip your hand in water... easy. Now slap the surface.... it's like hitting the table (almost). Pressures caused by breaking the surface make water act more solid on shorter timescales, which is why they say hitting water at high speeds is like hitting concrete; on those short times, it is actually like concrete!

Jim
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The ocean surface is not as hard as the ground but if you drop from a plane, you would hit it with such a high velocity that the pressure would most likely kill you or cause very serious damage.

Considering air resistance, the terminal velocity of a human, right before reaching the water, would be at most some $150\text{ m/s}$.

If you weigh $70\text{ kg}$, that would amount to a Kinetic Energy of

$$\frac12mv^2=0.5\times70\times150^2\text{ J}=787\ 500\text{ J}$$

Which is a LOT of energy, enough to crush many parts of your body even if you land on water. As ratchet streak mentioned, the water molecules can't move out of the way like they would do if you had fallen from a smaller height because of the high velocity. So you basically hit a semi-solid surface and all that energy comes back to you as a Reaction (Normal) Force.

JMCF125
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    Terminal velocity for a body falling is far less than 464.3 m/s. That's well over the speed of sound! – tpg2114 Apr 04 '14 at 15:18
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    Falling flat, a human body reaches ~50 m/s. Falling vertically, it's between 100-150 m/s. – tpg2114 Apr 04 '14 at 15:19
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    "The ocean floor" usually means the earth under the ocean, not the surface of the ocean. If you drop from a plane, you probably won't hit the ocean floor. – Joshua Taylor Apr 04 '14 at 15:21
  • @tpg2114 People here need to do their research before downvoting answers. http://hypertextbook.com/facts/JianHuang.shtml I am ignoring air drag so obviously it's not as fast. It's not a numerical question, I was just trying to show how fast someone falling from a plane can go. –  Apr 04 '14 at 15:42
  • @JoshuaTaylor The ocean floor mistake was stupid, I admit but it's just pathetic that you would downvote my answer because of that. –  Apr 04 '14 at 15:44
  • @ParthVader I agree; that would be a bad reason to downvote. Note that only have 101 rep here, not enough to downvote. It wasn't I that downvoted. – Joshua Taylor Apr 04 '14 at 15:53
  • The down-vote more likely was because of the ridiculously high value you got for terminal velocity. What about air resistance? – JMCF125 Apr 04 '14 at 15:54
  • @JMCF125 I was just giving an idea about how much the maximum terminal velocity of something falling from that height can be. I didn't say that you would fall at that height, that was self-explanatory, there are so many other factors to consider. Though tpg2114 is wrong in saying that it is not possible for a human body to reach that velocity, see the link above. –  Apr 04 '14 at 15:57
  • I also believe that is not possible, at least not in the troposphere. You said yourself: «I am ignoring air drag so obviously it's not as fast». «I was just trying to show how fast someone falling from a plane can go»: you just said they couldn't go so fast because of air drag! I think you should calculate only the kinetic energy, and assume some reasonable velocity, like the ones mentioned. – JMCF125 Apr 04 '14 at 16:06
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    I still don't see the point in calculating a velocity you know to be more than inaccurate. Why don't you just *assume a reasonable velocity*? – JMCF125 Apr 04 '14 at 16:11
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    But why add numbers at all if they are wrong? Just to show you know the equation for potential energy to kinetic? You have numbers for reasonable values of terminal velocity so why not use them? And the numbers close to what you have on the link you gave are marked as either unreliable or projected, and projected for falls from well over 11,000 m. – tpg2114 Apr 04 '14 at 16:11
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    Since your whole argument is "Your body has to absorb a huge amount of kinetic energy" then prove your point with an accurate amount of kinetic energy rather than one that's ridiculously high. You'll find your answer is just as valid but it is far easier to defend when you use real numbers. – tpg2114 Apr 04 '14 at 16:13
  • Well, tbh I had no idea what a reasonable value could be, so I thought of calculating it without the air drag. –  Apr 04 '14 at 16:16
  • Doing the calculations with air drag is also possible though. But the answer would get very long and I am too lazy to type all of that, especially in latex. –  Apr 04 '14 at 16:33
  • "If you drop from a plane, you probably won't hit the ocean floor." Unless you sink when you die. – Tim S. Apr 04 '14 at 17:04
  • If you jump out of a plane and go spread-eagle you can slow your speed to about 55 m/s: http://hypertextbook.com/facts/JianHuang.shtml. – Hot Licks Apr 06 '14 at 02:54
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    I think we're under-emphasizing how much kinetic energy that is. Let's try to put this in perspective: 787,500 J is enough energy to lift an F-150 pickup truck to the top of a 10 story building. That also means if you drop that pickup off a 10 story building into the water, the pickup would dissipate the same amount of energy as the human. And @user3058846 that means the impulse of force would be the same in those two situations. (And I'm thinking the pickup would hit the water pretty hard.) – Stainsor Apr 06 '14 at 04:59
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I'm not a physicist. So I am treading very carefully attempting to answer a question here... :)

A physical example that may help explain this is rock skipping. When you skip a rock, it will 'bounce' off of the water when at high speeds. Eventually it slows enough to no longer bounce but 'sink' into the water.

Picture your body doing the same thing. Your body will not want to sink into the water when going at that initial high speed as your body simply can't displace that water fast enough. So there's a force that acts back upon your body.

For a rock, not a huge deal. For a sack of living meat and blood and brains, it won't look pretty.

DA.
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    I think this isn't a great analogy because the vertical component of the velocity when a (skipping) rock hits the water is very small. Rock-skipping involves throwing the rocks horizontally so that they hit the water surface almost tangentially. – user80551 Apr 04 '14 at 18:35
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    @user80551 the analogy was meant to focus more on the forces that prevent the rock from just sinking. Yes, if you were 'skipped' across water you'd likely endure much less damage than 'dropped' but the concept is the same. – DA. Apr 04 '14 at 18:52
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When you go fast enough, the water molecules just can't move out of the way fast enough for a soft landing.

peterh
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ratchet freak
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