Answer
The magnitude of the impulse is $~~1760~kg~m/s$
Work Step by Step
In part (a), we found that the speed of the elevator after falling from a height of $36~m$ is $26.56~m/s$
If the passenger jumps, then the passenger's speed at the moment of impact is $~~(26.56~m/s-7.0~m/s)~~$ which is $~~19.56~m/s$
We can find the passenger's momentum at the moment of impact:
$p = m~v$
$p = (90~kg)(19.56~m/s)$
$p = 1760~kg~m/s$
The magnitude of the impulse is equal to the magnitude of the passenger's momentum.
Therefore, the magnitude of the impulse is $~~1760~kg~m/s$.
