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So me and my friend were having a discussion today that why should our legs break if we jump off from a taller height like 20 meters than if we just jump from a mere meter. We both agreed that our legs should break from a taller height, but we couldn't determine the reason behind it. I mean it doesn't matter if I drop from 20 meters or only 1 meter, the acceleration due to gravity will be same in both cases, right? So from Newton's second law, i.e. $F=ma$, my weight should be the same regardless of from which height I jump. Now, my weight is not gonna change even after hitting the ground, so shouldn't the reaction force that I get from the ground be equal to my weight since that is the force which I am applying on the ground during impact?

If no, then why not? Moreover, if the impact force is not the same as my weight (presumably higher than it), then why do I only apply my weight as a force on the ground when I stand on it, but the force becomes different on impact?

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The normal force from the ground must be large enough to give you an upward acceleration, that is large enough so that you don't melt into the ground. First of all, this means that the normal force will be time dependent: while you are in the air, the normal force is zero; then, as you hit the ground, the normal force grows very rapidly, causing acceleration so that you slow down; finally, the normal force reduces again so that when your velocity is zero, the normal force is equal to you weight, so that you remain on the ground.

When you are jumping from 1 meter, your velocity when you hit the ground is small, and the acceleration needed to make you stand still is small. Hence the maximum normal force does not need to be extremely large.

When you are jumping from 20 meters, your velocity when you hit the ground is large, and the acceleration needed to make you stand still is large. Hence the maximum normal force needs to be large*.

*Actually, the impulse must be large, which is the time integral of the normal force over the period of landing, from first contact to stand-still. This impulse is equal to the change in momentum.

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  • $\begingroup$ And how can I calculate the maximum reaction/normal force then since you mentioned that it increases to a max and then keeps on decreasing until it becomes equal to my weight. $\endgroup$ Dec 10, 2021 at 19:37
  • $\begingroup$ That is probably not possible without a lot of assumptions for a human. But if you model a human as, say, a spring obeying Hooke's law or something like that (we can be squeezed a little I guess), then you may be able to solve the equations of motion. But I don't know how to carry out that calculation other than in a computer program. $\endgroup$ Dec 10, 2021 at 19:39
  • $\begingroup$ I am guessing that the normal force will be the max as soon as the impact begins, and as the impact continues, the normal force becomes smaller and smaller until it becomes equal to my weight. So basically there will exist a net upward acceleration on me, being the highest at the beginning of impact and becomes zero as soon as my velocity becomes equal to zero. What would you say on this? $\endgroup$ Dec 10, 2021 at 19:48
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    $\begingroup$ You won't be able to calculate the maximum acceleration unless you have position/velocity data for many small time steps during the impact. But you can calculate the average acceleration very easily as long as you know the time of the impact. (You would need to measure this as well, but it is an easier measurement.) $\endgroup$
    – d_b
    Dec 10, 2021 at 20:02
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    $\begingroup$ The average force (magnitude) is $F_{\text{av}} = m \Delta v / \Delta t$, so the only influence the rigidity has is on the impact time $\Delta t$. More deformation during the impact will tend to increase the impact time, decreasing the average force. This is why it's better to land on an air bag or foam pad: because it increases the collision time. The effect of the rigidity on the maximum force will be extremely complicated for a realistic impact. $\endgroup$
    – d_b
    Dec 10, 2021 at 23:06
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If you jump from a height and hit the ground you are brought to a stop over a distance that is, roughly, equal to the distance between your feet and your centre of gravity. The force required to bring you to a stop is proportional to the square of your speed, so if, for example, you double your speed, you increase the force of impact by a factor of four.

Clearly if you jump from a greater height you increase your impact speed, and therefore you increase the force required to bring you to a stop.

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