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So I just started studying physics recently, and I came up with a question that I'm not sure how to solve, and I think it may be because of some fundamental misunderstanding. I don't want anyone to answer the problem, I'm just providing it for context.

The problem: Let's say a man is a distance h from the ground, and he has an indestructible box underneath him. He begins falling straight down, and once he's a distance d from the ground, he pushes straight down on the box. How much force must he push with to counteract the falling, reset his net velocity to 0, and land safely, assuming the force isn't great enough to shatter his skeleton. Also ignoring air resistance.

I figured this would be a trivial kinematics problem (maybe it is), but as I played with the equations I realized I had two conflicting intuitions:

  1. On one hand, there's only one constant force acting on him: gravity. If that's the case, shouldn't the force required to counteract the falling also remain constant (i.e. not dependent on the height from which he falls)?
  2. On the other hand, the amount of force one hits the ground with increases with fall height, so how is that force varying?

Clearly I'm missing something key here; any help is appreciated.

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  • $\begingroup$ what do you mean "reset net velocity to zero"? $\endgroup$ Sep 14 '21 at 1:44
  • $\begingroup$ As a hint, the force is related to the acceleration needed to bring the person to rest. The acceleration depends on the velocity (which you know), and the time interval over which the velocity decreases to zero. You may want to think more about this time interval. As a starting point, do you think a person wanting to minimize pain would prefer the time interval be large or small? $\endgroup$
    – Andrew
    Sep 14 '21 at 1:50
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The conceptual problem is balancing forces, when it requires balancing momenta. Of course:

$$ F = \frac{dp}{dt} $$

The gravitational force is indeed constant, so after falling for a time $t$:

$$ p = \int_0^t F(t')dt'=F\int_0^t dt'=Ft$$

He needs to apply a force to the box that gives him an impulse (the product for force times time) of $-p$.

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The reason why you are confused is that you are muddling a number of concepts.

Yes, gravity is the only force acting on the man while he is falling, so a force equal and opposite to gravity is sufficient to counteract the force of gravity. That happens when you are standing on the ground, for example, when the reaction of the ground prevents gravity from accelerating you downwards.

However, counteracting the force of gravity simply means that you stop any further acceleration, you do not bring the body to a halt if it is already moving downwards. A parachute acts in that way, providing a resistance that counteracts gravity completely, leaving the body to move downwards at a fixed speed.

However, stopping further acceleration downwards is an entirely different requirement from stopping the existing motion downwards- that is the root of your misconceptions. For that you need additional upward force.

The additional force required to reset the man's speed to zero depends upon how much time you are prepared to wait. Bungee jumping illustrates this point. The bungee rope stretches to the point at which it counteracts gravity and the jumper's speed stops increasing- thereafter, the rope continues to stretch and to apply an increasing force upwards to bring the jumper to a stop. By having a more or less elastic rope you can change the time, and therefore the distance, over which the jumper's fall is arrested.

You will see, therefore, that counteracting the force of gravity with an equal and opposite upward force is only sufficient to stop any further increase in downward speed- it does not reduce the downward speed.

Finally, if the man were to push on the box, then he could only do so for a moment by leaping upwards from it. Once he had lost contact he would continue to accelerate downwards at 1g. The effect of his jump would be in part to accelerate the box downwards, rather than propelling himself upwards. To restore his speed to zero, the momentary force he would have to apply with his legs to the box would be even greater than the force he would have to apply by simply hitting the ground- in either case, at any significant speed, the force would be massively greater than anything a human body could generate or withstand.

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It's important to note the distinction between impulsive forces as oppose to constant forces. An impulsive force is one that briefly acts on a body, changing its state of motion over a brief time, rather than a steady continuous force. A non-impulsive force is one like the gravitational force on an object, but an impulsive force can be one that happens during a collision with another object.

In this example, the force the man requires to stop himself (impulsive force which equals change in momentum needed at impact) increases with increasing height, and the force of gravity produces a constant acceleration $$g=\frac{F_g}{m}$$ where $F_g$ is the (constant) force of gravity.

If the man is in free-fall from a distance $h$ then at the instant when he hits the ground, he will have a velocity given by $$\frac{1}{2}mv^2=mgh$$ or $$v=\sqrt {2gh}$$ meaning that the height $h$ he falls from is what determines this final velocity. That is, although he is being accelerated by a constant force, the height will be what determines this speed.

If the man has mass $m$, then at the instant before impact, he has a momentum by $$p_b=m\sqrt {2gh}$$ This will require an impulse $$F\Delta t=m\sqrt {2gh}=p_a$$ where $\Delta t$ is how long it takes him to stop himself. Note that $$\vec p_b+\vec p_a=0$$ or $$\vec p_b=-\vec p_a$$ The impulse increases for increasing height, even though the gravitational force remains constant.

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The key here is momentum. The longer he's in freefall, the larger his downward momentum. After hitting the ground, his momentum will be 0 by some combination of pushing off the box and smacking the ground.

Say you can give yourself $10 \frac{kg \cdot m}{s}$ of upward momentum when you jump off a freefalling box. If your downward momentum is $100 \frac{kg \cdot m}{s}$ from falling, you'll still hit the ground with $90 \frac{kg \cdot m}{s}$ of momentum. How much momentum you hit the ground with (and how soft the surface is) determines the impact on your body.

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From the work/kinetic-energy theorem, the falling guy has to do enough work to counter his kinetic energy when he gets to the box. This leads to the equation

$W=\frac{1}{2}mv^2$

Using conservation of energy and assuming that the guy started falling from rest, it is seen that

$mgh=\frac{1}{2}mv^2$

where $h$ is his starting height above the box that he is going to land on. Since work is defined as the product of force and distance, this leads to the following:

$W=Fd=\frac{1}{2}mv^2=mgh$

where $d$ is the guy's stopping distance. Assuming that this stopping distance is constant (a good assumption), this leads immediately to the relationship between stopping force vs. falling height.

$Fd=mgh$

$F=\frac{mgh}{d}$

This means that the force that the guy has to apply to the box is directly proportional to the height that he fell from, because $m$, $g$, and $d$ are all constant.

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