We have the next formula
[tex]s=\sqrt[]{9.81d}[/tex]s is the speed in meter per second
d is the depth of the water in meters
If
s= 150m/s
d=?
so we need to isolate the d
[tex]s^2=9.81d[/tex][tex]d=\frac{s^2}{9.81}[/tex]we substitute the values
[tex]d=\frac{(150)^2}{9.81}=2293.57=2294m[/tex]The approximate depth of water where the disturbance took place is equal to 2294 meters