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Suppose in a situation there is a man falling from a height, when he falls from a height and touches the ground then he applies some force on the ground which is equal to his weight, and in another situation the same man with same weight is lying down on the ground.

So is it that that the force applied by the man on the ground will be same in both the cases?

I am not talking about gravitational force, I am talking about the other force being applied by the ball on earth and vice versa which balances the ball (please do not consider elasticity in this question).

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Answer of both case is different, in first case while the man is in air, he is in free fall with respect to earth, he apply force when he just touches he earth, there will be the impulse experience by the earth due to change in momentum of your body from $mv$ to $0$, While in the second the force is constant that is the weight of the body.

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No, the force applied in the two cases won't be the same.

The falling person will exert a force on contact to the earth which will be much higher than his weight. The force will be impulsive and is given by

$$\vec F=\lim_{\Delta t\to 0}\frac{\Delta \vec p}{\Delta t}$$

That's because in the process of falling the person has gained a lot of momentum and on contact with ground he will suffer a very large change in momentum in a very short amount of time (assuming the person is stopped very quickly) exerting a huge force on the ground and in turn he will also receive the same force in the upward direction by the earth and hence will suffer great damage.

(Please do not consider elasticity in this question )

The collision itself isn't elastic at all because the person won't bounce back if he falls on hard ground.

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No the force applied is not the same.

In the first case the force that is exerted on the earth is because of the exchange of momentum of body with the earth over a short instant of time.It means that when you fall and hit the ground the body will have zero velocity. It means you decelerate over a very short instant of time. This negative acceleration times the mass is the force that you exert on the earth.

In the second case,the force applied on the earth is your weight(the force with which the earth is pulling you).

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  • $\begingroup$ @gupta the force exerted by body in first case can be smaller than the weight of the body in case of highly deformable material $\endgroup$ – ASTRONO Oct 12 '19 at 13:28
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When a man falling it must carriyng some velocity before touching ground after touching ground its velocity change so change in momentum creating force on the ground.But the man who lying in ground not changing velocity so he is not applying any force.

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  • $\begingroup$ Thanks for helping. I think In the second case he is not changing his velocity because the forces are balanced out.Can you please give a more detail explanation of the second situation . $\endgroup$ – Yashvik gupta Oct 12 '19 at 14:55
  • $\begingroup$ Newton's third law $\endgroup$ – baponkar Oct 12 '19 at 14:58
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When the falling man hits the ground his impact force will be greater than his weight.

On way to determine the impact force is to apply the work energy theorem, which states that the net work done on an object equals its change in kinetic energy, or

$$Fd=\Delta KE$$

Where $F$ is the average impact force, and $d$ is the stopping distance.

Example: a 75 kg (165 lb) man steps off a ledge 2 meters from the ground and lands on his feet on a concrete surface. To account for some bending of the knees to ease the impact let his stopping distance be 0.5m. His kinetic energy upon impact will equal his loss of gravitational potential energy $mgh$. Neglecting the potential energy loss of the last 0.5 m we have

$F$ x 0.5 = 75 x 9.81 x 2

For an impact force on his feet of

$F$ = 2943 N= 662 lbf

Hope this helps

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  • $\begingroup$ Yes it helped a lot !! Thanks for the answer $\endgroup$ – Yashvik gupta Oct 12 '19 at 15:01
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It averages out.

All of the time that he's falling, he puts no force on the earth's surface. He is in free fall from the earth's force, and the earth is in free fall from his force.

Once he touches the earth, most of the kinetic energy he built up while he was falling converts to pressure on the earth's surface.


Edit:

Pressure is force per unit area. If he lies on the ground, the force that creates the pressure can be calculated. We could integrate that force over time, to get an average force.

We could do the same for the falling man. All the time he's falling, his pressure on the earth is zero. The force instead goes to increase his kinetic energy, and reduce his potential energy. His velocity increases linearly. Then when he hits the ground, the force is proportional to the deceleration. The amount of pressure-force integrated over time should equal the change in velocity over that time, which is the same change in velocity he got while he was falling, which is the same as the pressure-force for the man lying on the ground the whole time.

Or maybe I'm mistaken.

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  • $\begingroup$ What do you mean by averaging out ? $\endgroup$ – Yashvik gupta Oct 12 '19 at 14:49
  • $\begingroup$ I added more explanation. I hope I'm right. $\endgroup$ – J Thomas Oct 13 '19 at 17:45
  • $\begingroup$ Thanks for adding more explanation but I still didn't understand one thing, how is pressure in involved in this scenario ? I mean we are talking about the force exerted ,so what does pressure have to do with it ? $\endgroup$ – Yashvik gupta Oct 13 '19 at 17:49
  • $\begingroup$ And finally if I got you right you are saying the force will be same ? $\endgroup$ – Yashvik gupta Oct 13 '19 at 17:50
  • $\begingroup$ If you look at Newton, force only counts when it actually makes things move. When two forces precisely oppose each other there is no net force. But being subject to balanced forces is different from being free. If four horses are pulling you apart, we can say no work is being done until your arms are ripped off. But.... You asked about a man lying on the ground. He has no net force because he isn't moving. But there is pressure, which reflects the force of gravity. $\endgroup$ – J Thomas Oct 13 '19 at 18:42

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