Hello! 1. The problem statement, all variables and given/known data After the cable snaps and the safety system fails, an elevator cab free falls from a height of 36m. During the collision at the bottom of the elevator shaft, a 90 kg passenger is stopped in 0.005 s. Assume that neither the passenger nor the cab rebounds. What are the magnitudes of the: a) impulse and b) average force on the passenger during the collision? If the passenger were to jump upward with a speed of 7.0 m/s relative to the cab floor just before the cab his the bottom of the shaft, what are the magnitudes of the: c) impulse and d) average force (assuming the same stopping time)? 2. Relevant equations J = F⃗Δt= Δp⃗ = m(v⃗f−v⃗i) v2=v20+2aΔy 3. The attempt at a solution a) I used this formula v2=v20+2aΔy with y=-36m and v2=0 to solve for v1. I got that v1=26.56 m/s I then used J = Δp⃗ = m(v⃗f−v⃗i) using vf= 0 and v1=26.56 to get that J=-2390.7 kgm/s b) I used J = F⃗Δt with J=-2390.7 and t=0.005s to get that F= -478136.38 N (480000 N) c) I used J = Δp⃗ = m(v⃗f−v⃗i) again, but I was confused as to whether or not vi= 7 m/s (or if it was more complex than that). But I used vi=7 m/s to get that J = -630 kgm/s d) J = F⃗Δt with the new J to find that F=126000 N Am I on the right track? The numbers just seem so high...But I guess the dude's falling from quite a height. Thanks!