Answer
The magnitude of the impulse is $~~2390~kg~m/s$
Work Step by Step
We can find the speed after falling from a height of $36~m$:
$v^2 = v_0^2+2ay$
$v^2 = 0+2ay$
$v = \sqrt{2ay}$
$v = \sqrt{(2)(9.8~m/s^2)(36~m)}$
$v = 26.56~m/s$
We can find the passenger's momentum after falling from a height of $36~m$:
$p = m~v$
$p = (90~kg)(26.56~m/s)$
$p = 2390~kg~m/s$
The magnitude of the impulse is equal to the magnitude of the passenger's momentum.
Therefore, the magnitude of the impulse is $~~2390~kg~m/s$.
