What is the acceleration in meters per second^2 and g's during impact?

Question

Falls resulting in hip fractures are a major cause of injury and even death to the elderly. Typically, the hip's speed at impact is about 2.0 m/s. I this can be reduced to 1.3 m/s or less, the hip will not fracture. One way to do the is by wearing elastic hip pads. (a) if a typical pad is 5.0 cm thick and compresses by 2.0 cm during the impact of a fall, what acceleration in m/s^2 and g's) does the hip undergo to reduce its speed to 1.3 m/s? (B) The acceleration you found in part (a) may seem like a rather large acceleration, but to fully assess its effects on the hip, calculate how long it lasts.

Answer 1

One bit of data you are missing is the distance of the fall. I will assume 1 meter from the ground to the hip and that there is no resistance in the fall.

Using conservation of energy, the speed at impact is
v=sqrt(2*10*1)
v=4.472 m/s

In order to reduce the speed to 1.3 m/s in 2.0 cm, the following equation must be satisfied:

1.3=4.472-(a+9.81)*t
and
2.0/100=4.472*t-.5*(a+9.81)*t^...

the t is the time of compression and a is the deceleration.

The easiest way to solve is to isolate a*t from the first equation and substitute into the second


4.472-1.3+9.81*t=a*t

1/25=2*4.472*t-(4.472-1.3-+9.8...
solve the quadratic for t
and then solve for a
j