Speed and average force?

Question

A pole-vaulter of mass 63.7 kg vaults to a height of 6.01 m before dropping to thick padding placed below to cushion her fall. Find the speed with which she lands.

If the padding brings her to a stop in a time of 0.480 s, what is the average force on her body due to the padding during that time interval?

Answer 1

Considering ideal conditions that the she lands on her feet and not on any other body part, her legs provide no elastic force and the padding is at ground level.
Conserving energy,

Total energy at the top=Total energy at bottom
mgh=(1/2)mv^2
=>sqrt(2gh)
=>sqrt(2*9.8*6.01)=10.85 m/sec

F(avg)=change in momentum/Time

=>63.7(10.85 - 0)/0.48=1439.88 N

Answer 2

Ignoring friction and air resistance, we know the initial energy is equal to the final energy.

The great thing about energy is that we can take any two moments in time to get our answer. We're going to take the moment she reaches her peak and the moment she lands.

Initially, her speed is zero, so KE = 0. Finally, her height above ground is zero, so PE = 0.

Ei = Ef
PEi = KEi

m g h = ½m v²
g h = ½ v²
9.8 * 6.01 = 0.5 * v²
v = 10.853 m/s


For this, we need to know her acceleration.....

a = (v2 -v1)/(t2 - t1)

a = (0 - 10.853)/(0.480 - 0)
a = -22.6112

The a is negative because she is slowing down to a stop.

F = ma = 63.7 * -22.6112 = -1440.33 N

The force is negative because it is acting in the direction opposite her fall.... up.

Answer 3

Not possible to answer without knowing the level of the padding, the value of acceleration due to gravity (altitude) and the area of contact ie did she land on her head, back, butt etc. For true detail in figuring problems like this the best source I have found is the Snipers Handbook. It brings up questions and issues that you would never think of.