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Say you have two situations with an object interacting with the ground. In the first situation the object has been sitting on the ground for a while so the force will be mg. In the second situation the object has been falling for a while and finally hits the ground, will the force exerted on the ground still be mg?

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When the object is sitting on the ground, yes, the normal reaction force (physics jargon for upwardforce exerted by ground on object/object on ground) is $mg$.

A falling object is much more interesting. Since it was falling, it had a velocity. This velocity is gone now, so we had a deceleration.

Just for the sake of simplicity, let's say the deceleration was constant ($a$, positive value pointing upwards). Here's our FBD:

enter image description here

We get the equation $$ma=N-mg \implies N=ma+mg$$

So, the force exerted is augmented.

With most bodies, the deceleration is almost instantaneous. Since $v=u+at$,$v=0$, and $t=\rm small$, we get that $a$ is large. For a more general case, considering the fact that acceleration varies, we can use our impulse equation $\int N dt=m\Delta v$, and this also shows that $N$ is large if $t$ is small.

So, the "extra" force is huge. What happens is that you get an extremely large force in an extremely short time. The net effect isn't that noticeable--even though the force, if applied for a longer time, can easily destroy stuff. Though if you see professional baseballers/cricketers, they always bring their hands down with the ball--this increases the time and decreases the force--since a pop fly still hurts and has an unimaginable magnitude of force exerted.

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