Physics 101 (free fall impact)

[TransformX]

I know I've asked it before and I remember I did not feel satisfied with the answer, so here I go again, but this time I hope I am clearer, just to be sure I understood things well:

I am basing my assumptions on: http://www.engineeringtoolbox.com/im...ce-d_1780.html


1 kg = 9.8 Newton

Free fall: F = m g h / s
m: Mass
g: Gravity = 9.81 m/s²
h: Height
s: stopping distance

So dropping a 2kg brick from 2 meters high onto a concrete floor (assuming a stopping distance of 1mm):
F = 2 * 9.81 * 2 / 0.01 = 3924kN

3924kN / 9.8 (Newton per kg) = 400kg (weight as moment of impact)

[cjolley]

Keep in mind that the stopping distance would be less than that and the resulting force higher against a concrete floor. And since the brick will be crushed a bit or broken the whole calculation will come unglued. Better to use a steel ball and metal floor as an example.

Though, my memory of that calculation used the amount of time the the impact lasted, not distance, but I guess either could be used. It's really just a way of calculating the force it takes to decelerate the brick in a given distance (or time).

It's the short stopping distance (or time) that makes the force so high.
think of the difference between driving in a nail or trying to push the nail in with the hammer head.