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?
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:
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.