Answer
The average deceleration while under the water is $20 ~m/s^2$.
Work Step by Step
In the air:
$v^2 = 2ay$
$v = \sqrt{2ay} = \sqrt{(2)(9.8)(4.0)} = 8.9 ~m/s$
In the water:
$a = \frac{v^2-v_0^2}{2y} = \frac{0-(8.9)^2}{(2)(2.0)} = -20 ~m/s^2$
The average deceleration while under the water is $20 ~m/s^2$.
