[eugenebrooks]

The prior post is correct. The answer can be more easily
understood from the point of view of conservation of energy.

The total change in gravitational potential is m g h, where h is
the distance of the drop, 3.81 meters.

To get the average force, f, applied during the stop equate this total
energy to force times, f, times the distance of the stop, 0.01 meters.

50 * 9.80 * 3.81 = f * 0.01

Now, f = m a, so we have

50 * 9.8 * 3.81 = 50 * a * 0.01

And finally, putting a in terms of gees, a factor times the
acceleration of gravity,

50 * 9.8 * 3.81 = 50 * gees * 9.8 * 0.01

We see that the mass cancels, and the acceleration of gravity
cancels, and the number of gees is 3.81/0.01 = 381

At this point, you should see that it is not hard to go
a little further and account for the gravitational potential
for that last centimter of crunch into the concrete, and
then get into some deep thought about the gee of gravity
and when it kicks in, so to speak.

Eugene