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.

  • 2
    $\begingroup$ Related: physics.stackexchange.com/q/106826/29216 $\endgroup$ – BMS Apr 4 '14 at 16:27
  • 4
    $\begingroup$ 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! $\endgroup$ – Renae Lider Apr 4 '14 at 17:42
  • 46
    $\begingroup$ Including the experimental-physics tag on this question seems rather morbid. $\endgroup$ – jpmc26 Apr 4 '14 at 19:14
  • 3
    $\begingroup$ Being unconscious and severely wounded on land is bad. Being unconscious and severely wounded on the open ocean really sucks though. $\endgroup$ – Christian Apr 4 '14 at 21:14
  • 2
    $\begingroup$ @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: en.wikipedia.org/wiki/Vesna_Vulovi%C4%87 veteranstoday.com/2011/05/09/… $\endgroup$ – Christian Apr 4 '14 at 22:05

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?)

  • $\begingroup$ 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 $\endgroup$ – Jim Apr 4 '14 at 14:52
  • 1
    $\begingroup$ 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]. $\endgroup$ – Volker Siegel Apr 4 '14 at 15:01
  • $\begingroup$ 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. $\endgroup$ – Volker Siegel Apr 4 '14 at 15:06
  • 2
    $\begingroup$ 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 $\endgroup$ – Jim Apr 4 '14 at 15:26
  • 2
    $\begingroup$ 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. $\endgroup$ – André Apr 8 '14 at 14:49

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.

  • 2
    $\begingroup$ 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. $\endgroup$ – Hagen von Eitzen Apr 5 '14 at 18:18
  • 6
    $\begingroup$ @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. $\endgroup$ – Taemyr Jun 11 '15 at 8:21

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!


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.

  • 5
    $\begingroup$ Terminal velocity for a body falling is far less than 464.3 m/s. That's well over the speed of sound! $\endgroup$ – tpg2114 Apr 4 '14 at 15:18
  • 4
    $\begingroup$ Falling flat, a human body reaches ~50 m/s. Falling vertically, it's between 100-150 m/s. $\endgroup$ – tpg2114 Apr 4 '14 at 15:19
  • 8
    $\begingroup$ "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. $\endgroup$ – Joshua Taylor Apr 4 '14 at 15:21
  • 4
    $\begingroup$ 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? $\endgroup$ – JMCF125 Apr 4 '14 at 16:11
  • 4
    $\begingroup$ 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. $\endgroup$ – tpg2114 Apr 4 '14 at 16:11

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.

  • 4
    $\begingroup$ 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. $\endgroup$ – user80551 Apr 4 '14 at 18:35
  • 1
    $\begingroup$ @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. $\endgroup$ – DA. Apr 4 '14 at 18:52

When you go fast enough, the water molecules just can't move out of the way fast enough for a soft landing.


protected by Qmechanic Apr 4 '14 at 18:17

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.