You are coordinating a stunt for a movie. A stunt person with a mass of 60 kg must fall a distance of 32 m and land on a large, inflated cushion. In order to avoid any serious injury, you want to make sure that the performer does not experience an average force of larger than 3000 N. What is the shortest period of time over which the cushion can break their fall?

First of all, we have to calculate the final velocity of the person as it hits the cushion. Since the kinetic energy gained during the fall is equal to the gravitational potential energy lost, we can write

[tex]\frac{1}{2}mv^2 = mg\Delta h[/tex]

where

m = 60 kg is the mass of the person

g = 9.8 m/s^2 is the acceleration of gravity

v is the velocity at the moment of impact

[tex]\Delta h = 32 m[/tex] is the change in heigth

Solving for v,

[tex]v=\sqrt{2g\Delta h}=\sqrt{2(9.8)(32)}=25.0 m/s[/tex]

After landing on the cushion, the man is brought to rest, so its change in velocity will be (in magnitude)

[tex]\Delta v = 32 - 0 = 32 m/s[/tex]

We can now use the impulse theorem to find the shortest duration of the collision such that the maximum force is F = 3000 N:

[tex]m \Delta v = F \Delta t[/tex]

where [tex]\Delta t[/tex] is the shortest period of time. Solving for it,

[tex]\Delta t = \frac{m \Delta v}{F}=\frac{(60)(32)}{3000}=0.64 s[/tex]