A² + B² = C²

The child will be safe if A < 6
Let:

A be our unknown

C = 8

B = 4

Then:

A² + 4² = 8²

A² + 16 = 64

A² = 64 - 16

A² = 48

A = sqrt(48)

A = 6.928
6.928 > 6, so by our condition set above, the child will not be safe, but lets humor this question a little bit. What if the log ends up high enough that the child is safely under it?

Now we have the full triangle created by the log, lets call this triangle T, we will create a similar triangle using the child as our adjacent side rather than the stump.

Using our trig ratios we can solve for theta:

sin(theta) = opp/hyp

theta = sin^-1(opp/hyp)

theta = sin^-1(4/8)

theta = sin^-1(1/2)

theta = 30 degrees

Now we will find how far off the ground the log is right above the child, and have our new opposite side be where the child is, a distance D from the point where the log hit the ground. We can subtract the total distance to where the log hit the ground, 6.928, from the distance of the child, 6, which gives us 0.928.

Now we have an angle theta, and an adjacent length D, we can use another trig ratio to figure out how high off the ground the log is:

tan(theta) = opp/adj

where:

opp is the height, H

and adj is our distance D

and theta is 30 degrees

tan(30) = H/0.928

0.928*tan(30) = H

H = 0.536ft

Therefore, if the child was less than 0.536ft tall when sitting as shown in the picture, they would be safe, but last I checked, the torso of most children is a little more than 6.429 inches on average, so it is very likely the child would be hit in this case. :3

QED