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If a person jumps from the first floor of a building and lands on a concrete surface, they will suffer serious injury because of Newton's third law.

If the same person jumps the same distance and lands in swimming pool filled with water, however, then there will not be any serious injury.

The person in both cases lands with same amount of force. Why doesn't water offer the same amount of force in return as concrete?

muru
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Sim Cup
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    It depends on matter phase. Newton laws assumes interacting bodies are in solid state, otherwise if one of them - are not,- this changes game rules. Similarly or even worse,- if you jump into gas chamber,- you won't notice at all that gas has "punched you", unless it is frozen into solid. For gas effect on you- you need to apply ideal gas law / statistical physics and such, and not Newtonian classical mechanics for solids. – Agnius Vasiliauskas Aug 23 '22 at 15:39
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    Related: https://physics.stackexchange.com/q/9059/ – hft Aug 23 '22 at 16:38
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    @AgniusVasiliauskas I don't see what you're getting at - Newton's laws apply to any matter, regardless of phase. It's merely convenient to apply them to a system of rigid point particles, but it's not like they don't apply to liquids or gases. A fish swimming through water or a plane flying through the air can both be modeled with Newton's laws - a fish/plane is propelled forward because it pushes fluid backward. No assumption of matter phase is needed to apply Newton's laws. – Nuclear Hoagie Aug 23 '22 at 16:40
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    @NuclearHoagie I'm not saying that Newton laws does not apply to gas or liquids. What I say is that for determining exact effect of gases/liquids to body immersed in them,- you need Newton mechanics applied to the system of molecules, i.e. you need idea/real gas law or Navier-Stokes equations or at least statistical physics. They all use Newtonian laws/mechanics to some degree, but it's another level. You don't calculate each molecule movement for determining your car speed, right ? Similarly, it's not wise to directly apply $10N$ force to the air and see what reaction force will come out. – Agnius Vasiliauskas Aug 23 '22 at 17:38
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    Pinning the injuries to the 3rd law is like shooting the messenger. The 3rd law just states whatever happens to one body, it also happens in equal and opposite measure to the other body. The 3rd law is just the mechanism where energy is transferred, but what you are after is how is the energy generated and absorbed, which is actually the 2nd law. – John Alexiou Aug 23 '22 at 22:01
  • @JohnAlexiou How is the way of energy generation and it's absorption related to 2nd law? I just know force equals ma.Sadly, I don't know it from energy perspective.Please explain and Do suggest a good book on this topic also. – Sim Cup Aug 24 '22 at 00:15
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    See also https://physics.stackexchange.com/q/718786/ – Rmano Aug 24 '22 at 10:46
  • "differ in when" is a readability speed bump. :/ (you don't need the word 'in') – Wyck Aug 24 '22 at 14:25
  • Related question: https://physics.stackexchange.com/q/470714/179151 – BioPhysicist Aug 25 '22 at 14:53
  • "they will suffer serious injury": very unlikely for a jump from the first floor! Maybe a sprained ankle if you're not careful. – TonyK Aug 25 '22 at 15:45

9 Answers9

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It is not the case that you "land with the same amount of force" - you land with the same amount of kinetic energy, the difference is how long it takes to dissipate that energy. It all comes down to the "stopping time" - when you land on concrete, you go from your impact velocity to zero velocity in a fraction of a second. When you land in water, you plunge below the surface and come to a stop quite a bit slower, over the course of many fractions of a second.

$F=ma$, and $a = \Delta v/\Delta t$. In both cases, $\Delta v$ is the same (you go from impact velocity to 0), but when you land in water, $\Delta t$ is much greater, making $a$ and therefore $F$ much lower. This is the same principle behind crumple zones in cars, or why you should bend your knees when landing a jump - by extending the deceleration time, you decrease the force exerted.

The reason why the deceleration times are different between concrete and water is related to the fact that concrete is a solid and water is a liquid. The molecules in concrete are locked into a rigid configuration. Concrete molecules don't move around freely - when you push on concrete, the concrete doesn't move, it pushes back to resist even large forces. Molecules in water, on the other hand, freely flow past one another - when you push on water, it accelerates out of the way. When confronted with a large force, a material can either resist it (like concrete), or yield to it (like water). Imagine being on ice skates - you can push off a rigid wall to accelerate yourself backwards, but if you push off another person on skates, you won't move as quickly, since the thing you're pushing off of yielded to the force of the push.

  • If we consider the case of person falling into water,same as big sphere colliding with very little small stationary spheres (like water molecules) then the force with which big sphere hits the first layer of small sphere should be the same force experienced by the big sphere.I have seen examples where the big sphere collide with small sphere and both move in same direction after collision which explains why person continue to go through water.But after collision as per Newton's law,person should experience the same force as it gave to first layer of water just like concrete but it doesn't.Why? – Sim Cup Aug 23 '22 at 16:52
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    @SimCup Added a bit more about the material properties. Basically, water molecules are free to move out of the way, while concrete molecules are not. With water, you can dissipate the force by moving water out of the way "bit by bit" (but quickly). When landing on a rigid surface like concrete, there is no "first layer" of concrete, the entire mass of the landing pad affects you through its rigidity. – Nuclear Hoagie Aug 23 '22 at 17:14
  • Even though the water molecules (or a Small affected volume of water) are free to move,the movement has happened because of collision.My confusion is that in any collision between two things ,both the things will experience same force,then why the person hitting the individual parts of water (or molecules) with some force, do not experience the same force themselves at that instant when it makes contact with the water? – Sim Cup Aug 23 '22 at 17:49
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    @SimCup The water experiences an equal and opposite force as the person landing in water. The concrete also experiences an equal and opposite force as the person landing on concrete. The forces experienced by the water and concrete are not the same, though. I think your mistake is thinking that the person hits whatever they're landing on with some fixed force, that's not the case - the force exerted upon landing depends on the material. Only the energy of the impact is the same - the force that the concrete pushes back with is much higher due to its rigidity. – Nuclear Hoagie Aug 23 '22 at 18:08
  • Do correct me if i get this wrong : When any object collide with other object then the force experienced by the first object totally depends on the resistance that the other object can offer.If it cannot offer the resistance and just moves then the force with which the first object is moving (in our case gravity) is irrelevant to the force experienced during the collision between two objects. – Sim Cup Aug 23 '22 at 18:18
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    @SimCup That's basically correct. Think of the old cinema trope where someone falls off a tall building, rips through a bunch of window awnings on the way down, and lands safely on the sidewalk. The window awnings can't provide any more resistance than they're structurally able to, they provide some resisting force and then give way. Landing in water is similar, the water provides some resisting force but mostly just gets out of the way - there's no structural rigidity, just viscous forces holding the water together, which can't push back very hard. – Nuclear Hoagie Aug 23 '22 at 18:28
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    @SimCup Similar to the difference between crashing a car into a brick wall, versus driving through a lightweight picket fence. The fence can't exert as much force before breaking/getting out of the way, so it results in a less serious crash. Driving a car normally is constantly colliding with the air in front of the car, but air gets out of the way so easily we don't even think about it. – Nuclear Hoagie Aug 23 '22 at 18:31
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    it may be worth stating explicitly that the definition of a (used for F=ma) here is specifically the average accelaration, and hence the average force. For the purposes of comparing impact on different materials this can be a useful proxy for the more relevant quantity, the peak force, but unless the acceleration is constant (which it certainly won't be for an elastic collision with concrete) the two are not the same – Tristan Aug 24 '22 at 12:52
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    Considering $dp=Fdt$ and $dE=Fds$, in the first paragraph where you mention kinetic energy, I would say "stopping distance", not "stopping time". The second paragraph can be written in terms of momentum, so here "stopping time" fits. This nitpicking does not change the conclusion and I upvoted your answer. Still, since the question is about force and you introduce kinetic energy, "stopping distance" seems more natural to me in this context, because the direct relation between the force and the kinetic energy is by the distance. Xor change to "you land with the same amount of momentum". – Kamil Maciorowski Aug 24 '22 at 21:18
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    @SimCup If it makes it easier for you to imagine consider a person falling from the same height and landing on air (this can actually happen, someone jumping off a plane in a skydive must first fall the distance you mention and land on nothing/air). Imagine yourself falling onto air, you don't feel anything "hit" you because you feel that thing blowing around you instead. The air is loose enough to move out of the way. – slebetman Aug 25 '22 at 07:58
  • Hence the old saying: "It's not the V_f = V_i + a t that kills you, it's the F = m ΔV / ΔT." – Chris Bouchard Aug 26 '22 at 01:10
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But the person in both case lands with same amount of force. Then why doesn't water offer the same amount of force in return as concrete does?

This is not correct: the force that the person applies to concrete/water is the same as the force that the concrete/water applies to the person (Newton's third law). The force is different in the two cases, as the person is slowered down for different amount of time, i.e., their acceleration is different: when they fall on concrete, they are stopped almost immediately, the force is high (Newton's second law), whereas when falling in water they are slowered down gradually, while submerging to a noticeable length - the force is smaller.

Roger V.
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It is not the impact itself, but how fast you come to a stop that dictates the overall effect felt by your body. At low velocities (the first floor case, for instance), the water is able to "get out of the way" and hence slows the body down gradually and not with a sudden jerk.

The force applied by the human on the concrete and vice versa, and the force applied by the human on the water and vice versa are equal in both cases because of the third law, but it is the rigidity of concrete that makes the body come to a sudden stop.

At higher speeds, however, (a tenth floor jump, for example) the water simply cannot "get out of the way" fast enough, and the impact felt would be the same as hitting concrete.

Hritik Narayan
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Newtons third law says that the $F = ma$, where the acceleration $a$ is the change of velocity per time $\Delta v/ \Delta t$. At the instant you hit the ground, $\Delta v$ is very high if you are falling fast, resulting in a high force. When you fall on the water, $\Delta v$ is less, since you do not totally stop. The reason you are confused is because just before you hit the ground/water, the force is indeed the same, but exactly when you hit the surface, it is different for both cases.

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But the person in both case lands with same amount of force. Then why doesn't water offer the same amount of force in return as concrete does?

The person does not land with the same amount of impact force. The average impact force that the concrete exerts on the person is greater than the average impact force the water exerts on the person because the person's stopping distance is much less for the concrete.

This can be seen by applying the work energy principle, which states that the net work done on an object equals its change in kinetic energy, along with the definition of work.

The work done on the body by the concrete or water where $F_{ave}$ is the average impact force and $d$ is the stopping distance

$$W=F_{ave}d$$

Ignoring the change in gravitational potential energy after impact, this work equals the change in kinetic energy of the object that is brought to a stop, or

$$F_{ave}d=-\frac{1}{2}mv^2$$

Where $v$ is the velocity of the person just prior to impact and the final velocity is zero when brought to a stop. So

$$F_{ave}=-\frac{1}{2d}mv^2$$

Since the concrete gives very little compared to the water, the stopping distance $d$ for the concrete is much less than the water, meaning the average impact force (and the damage it does) is much greater for the concrete.

Hope this helps.

Bob D
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Lets look at the energy conservation

$$\frac{m}{2}\,v_i^2+m\,g\,x_i=\frac{m}{2}\,v_{f}^2+F_{f}\,x_{f}$$

where f is the final state ans i is the initial state

if both case is the final velocity $~v_f=0~$ but the distance $~x_{fc} \ll x_{fw} $ this means that the force that injured you $F_{fc} \gg F_{fw}$

where "c" for concrete and "w" for water

Eli
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We should calculate the Force required to break the concrete. I don’t think that the force which is generated from falling certain height is enough to break the concrete since it’s Mechanical properties are strong enough to withstand, that’s what we say resistance by the solid body when you apply force on it on some particular area. Same if you compare to the water it’s mechanical properties are very very small compared to solid and it is not able to resist that falling body force, it shear off and it couldn’t able to resist like a solid. If we think this way we may get some answer.

Thanking You

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Water can not with stand shear strain obviously. Therefore much more of the surrounding material can contribute to the mass in the opposing force then can water.

marshal craft
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As someone more creative than I once pointed out, we think of gravity as being a powerful force, but these things are relative. Gravity (1G) will accelerate you from rest to 100kph in the matter of a few seconds over a few floors drop. But the atomic forces in the concrete will decelerate you from 100kph to zero in a few milliseconds over a distance of a few microns, so that you experience many tens to many hundreds of Gs of force. That's what we mean by a resistant material.

Some of your respondents have overstated the resistance effect of water in decelerating the body from great height and hence great impact velocity. It depends. There are cases of people surviving jumping off the helideck of a burning oil rig, and of those that didn't make it some will have drowned because they didn't rise back to the surface in time. Bolt upright feet first entry is probably best it seems.