Ok.

Sorry, your units were unclear, so I've made a few estimates/assumptions.

Force = mass x acceleration.

where acceleration is a change in velocity over time or distance.

Wikipedia has a bunch of equations about velocity and acceleration:The acceleration due to gravity, G is 9.8 [tex]\frac{m}{s^}[/tex]

So, if a person falls for one second of free-fall

one reaches a final velocity of 9.8 m/s, and travels 4.9 m.

Now, if you slam your head into concrete travelling at 9.8 m/s, you change from 9.8 m/s to 0 m/s in a certain distance.

Your body will be able to absorb the impact over a few cm, depending on how you hit, and if you are rolling, and perhaps on how much "extra padding" you have.

However, the head is far less forgiving. Perhaps being generous, one might consider the deceleration to occur in ½ cm, or 0.005 m.

Assuming constant acceleration.

Δv = a*t (change in velocity = acceleration * t)

Δv = [tex]\sqrt{2a*d}[/tex] (change in velocity = square root of 2 * acceleration * distance)

t = [tex]\frac{d}{\frac{v}{2}}[/tex] = [tex]\frac{2d}{v}[/tex] ( time = distance / average velocity)

Ok, let's take the second equation, and solve it for acceleration.

Δv = [tex]\sqrt{2a*d}[/tex]

Δv

^{2} = 2a*d

a = [tex]\frac{\Delta v^2}{2d}[/tex]

So, substituting the values we have:

a = [tex]\frac{(9.8\frac{m}{s})^2}{2*0.005 m}[/tex] = 96,040 [tex]\frac{m}{s^2}[/tex] = 9,800G.

Ok, so force = mass x acceleration (F=ma)... and I could go ahead and calculate the force in Newtons. However, now that I have the acceleration expressed in G's, one can just multiply it by the pounds.

Now, in most cases, the head won't take the full weight of the child. So, why don't we use 3 kilos, or 7 lbs for the weight of a child's head. Let's also say the impact area is 5" x 5", or 25 square inches.

So, at 7 lbs for 1 G, we get 7*9,800 = 68,600 pounds of force.

Divide that by the area of impact to the head 25 sq inches, and we get 2,744 PSI.

Now, other than the time and distance I used to estimate the height of the fall (which you can put in your values), I have an estimate of the distance for the deceleration (½ cm). It all depends on the actual pressure, deflection, where it hit, and etc. I also estimated 25 square inches for the impact area, but it would depend on what part of the head hit. The nose certainly covers less area.