Jumping up before the elevator hits. after the cable snaps and the safety system fails, an elevator cab free-falls from a height of 36 m. during the collision at the bottom of the elevator shaft, a 90 kg passenger is stopped in 5.0 ms. (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

There are two sections of solution to this problem. The first is the impulse and the second is the force.

A.) In physics, when two objects collide, there is a small interval of time when these objects are in contact with each other. The net force applied on the two objects as one system at that time is called the impulse. Its equation is

Impulse = 2mv/t, where m is the total mass of the system, v is the velocity at impact and t is the time when the objects are in contact

But first, we have to find the velocity of impact. For free-falling objects, there is a derived equation for the velocity of impact: v = √2gh, where g is equal to 9.81 m/s^2 and h is the height of fall. Thus,

v = √2(9.81)(36) = 26.58 m/s
Impulse = 2(90 kg)(26.58 m/s)/(5×10^-3 seconds)
Impulse = 956,880 Newtons

B.) According to Newton's second law of motion: F=ma, where F is the net force applied on the system, m is the mass and a is the acceleration. For free-falling objects, the acceleration is due to gravity which is equal to g=9.81 m/s^2. Thus,

F = (90kg)(9.81 m/s^2)
F = 882.9 Newtons

It's because 3 if it was 10 it wouldn't look right do it goes up to 4d