Why does the opposing force differ in when falling on concrete vs on water in spite of Newton's third law? 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?
 A: 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.
A: 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.
A: 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.
A: 
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.
A: 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
A: 
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.
A: 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
A: 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.
A: 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.
