Answer
The magnitude of the impulse is $~~1074~kg~m/s$
Work Step by Step
We can find the speed after falling a total of $12~m$:
$v^2 = v_0^2+2ay$
$v^2 = 0+2ay$
$v = \sqrt{2ay}$
$v = \sqrt{(2)(9.8~m/s^2)(12~m)}$
$v = 15.34~m/s$
If we estimate the person's mass as $70~kg$, we can find the momentum when he reaches the surface of the water:
$p = (70~kg)(15.34~m/s)$
$p = 1074~kg~m/s$
In order to bring the person to rest, the impulse from the water must be equal in magnitude to the person's momentum.
Therefore, the magnitude of the impulse is $~~1074~kg~m/s$.
