It's hard to think of a physical system involving a force that acted for zero time. However I think it's useful to consider a collision, perhaps between two billiard balls.
When the balls collide they change momentum. We know that the change of momentum is just the impulse, and we know that the impulse is given by:
$$ J = \int F(t)\,dt $$
where I've used an integral because the force is generally not be constant during the collision.
If we use soft squidgy balls then the collision will take a relatively long time as the balls touch, then compress each other, then separate again. If we use extremely hard balls the collision will take a much shorter time because the balls don't deform as much. With the soft balls we get a low force for a long time, with the hard balls we get a high force for a short time, but in both cases (assuming the collision is elastic) the impulse (and change of momentum) is the same.
When we (i.e. undergraduates) are calculating how the balls recoil we generally simplify the system and assume that the collision takes zero time. In this case we get the unphysical situation where the force is infinite but acts for zero time, but we don't care because we recognise it as the limiting case of increasing force for decreasing duraction and we know the impulse remains constant as we take this limit.
I'm not sure it's helpful to think about the gravitational force, because I can't see a similar physical system where we can imagine the gravitational force deliverting a non-zero impulse in zero time.
Response to edit:
In you edit you added:
If I got it right, you are saying that we must consider it impulse when t=0?, else it is force.
I am saying that if we use an idealised model where we take the limit of zero collision time the impulse remains a well defined quantity when the force does not.
However I must emphasise that this is an ideal never achieved in the real world. In the real collisions the force and impulse both remain well behaved functions of time and we can do our calculations using the force or using the impulse. We normally choose whichever is most convenient.
I think Mister Mystère offers another good example. If you're flying a spacecraft you might want to fire your rocket motor on a low setting for a long time or at maximum for a short time. In either case what you're normally trying to do is change your momentum, i.e. impulse, by a preset amount and it doesn't matter much how you fire your rockets as long as the impulse reaches the required value.
Response to response to edit:
I'm not sure I fully grasp what you mean regarding the book, but the force of gravity acting on the book does indeed produce an impulse. Suppose we drop the book and it falls for a time $t$. The force on the book is $mg$ so the impulse is:
$$ J = mgt $$
To see that this really is equal to the change in momentum we use the SUVAT equation:
$$ v = u + at $$
In this case we drop the book from rest so $u = 0$, and the acceleration $a$ is just the gravitational acceleration $g$, so after a time $t$ the velocity is:
$$ v = gt $$
Since the initial momentum was zero the change in momentum is $mv$ or:
$$ \Delta p = mgt $$
Which is exactly what we got when we calculated the impulse so $J = \Delta p$ as we expect.
I think impact should depend on Amount of force applied by object? Or momentum?
Yes, if by impact you mean force, then the force experienced by the object that is hit depends on the momentum of the colliding object.
There is a formula from Newton's 2nd law, Force x time = change in momentum. If the colliding object is brought to rest $$Ft=mv$$ or $$F = \frac{mv}{t}$$
For the waterfall or falling water example, let's consider the following.
Pouring water from a jug at constant rate (kilograms per second) and stopping it with your other hand at different heights.
The $\frac{m}{t}$ part of the equation is the same, but the velocity of a falling object increases as it falls. Your hand experiences more force from the water when it's further underneath the jug simply due to the $v$ being greater further down.
Best Answer
You're looking at the net force acting on him at the instant he begins falling. And you are correct that, in the absence of air resistance, the net force that is acting on the man the instant after he begins falling in the same, given by $F_g=mg$. However, there's an old saying that says, "it's not the fall that kills you; it's the sudden stop at the end." This couldn't be more true, so to understand the difference between the two falls, we need to look at the force acting on the man as he's hitting the ground.
One good approximation to look at this is through the idea of impulse. Impulse is defined as the change in momentum of an object. At the instant before the man strikes the ground in each scenario, he has some momentum, $p$, given by $p=mv$ where $m$ is the mass of the man (in your example, 50 kg) and $v$ in the instantaneous velocity of the man right before he strikes the ground. At this point I'll do a little bit of hand-waving, and just say that the velocity right before striking the ground will increase as the height increases. Therefore, the $v$ before striking the ground for the 50m drop will be much higher than the $v$ from the 2m height. For the purpose of this problem, exact velocities are not important.
Ok, so now let's have the guy actually hit the ground. Although you may think that this occurs instantaneously, it actually occurs over a non-zero time interval. This is easily seen if you ever watch a collision through the aid of slow-motion. Anyway, so when the guy hits the ground, it takes some non-zero time interval, $\Delta t$ to come to rest. Assuming the guy is hitting the same surface in both scenarios, we'll assume $\Delta t$ is a constant across our trials. (This isn't quite true, but slight variations in $\Delta t$ across drop heights are negligible compared to wildly different impact velocities.)
When the guy hit's the ground, his momentum quickly goes from $p$, which is said is equivalent to the expression $mv$ to zero. This is because after he's fully collided with the ground, he's not moving anymore so his velocity, and therefore his momentum is zero. There's an equation that can be used to express this change in momentum, or impulse, in terms of force, and it's given by $$F\Delta t=m\Delta v$$ If the final velocity is zero, then $\Delta v$ is just given by $-v$, where $v$ is the velocity just prior to striking the ground. Solved for the impact force, $F$, we get that $$F=\frac{mv}{\Delta t}$$
Since the mass of the man, $m$, is the same no matter what height he is dropped from, and I said that we can (for our purposes) treat the time interval $\Delta t$ as a constant as well, you can see that the impact force, $F$, is proportional to the velocity $v$ the jumper has when he hits the ground. And as a mentioned earlier, higher drop height means higher final velocity, and therefore larger impact force.