Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.37 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 2.1 mm. If the floor is carpeted, this stopping distance is increased to about 1.0 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume that the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate.

Answer 1

Calculaqte the speed of the head hitting the floor using
V = sqrt (2gH) , where H = 0.37 m

Let the stopping distance be X.

The deceleration rate required to stop the fall is

a = V^2/(2X) = 2gH/2X = 2 g H/X

The duration of deceleration is

t = X/(V/2)= 2H/V

since V/2 is the average velcoty during deceleration.

Use those formulas to come up with the numbers you need for the two floor coverings.