What is happening here is that you have a question where
drag enters into the equation. If I drop a bowling ball and a BB, they both fall at the same rate as
drag, or in this case wind resistance, is negligible. As water is 600 times denser than air, you have to factor in
drag.
Here's the equation:
Fd is the force of drag,
p is the density of the fluid,[3]
v is the speed of the object relative to the fluid,
A is the reference area,
Cd is the drag coefficient (a dimensionless parameter, e.g. 0.25 to 0.45 for a car), and
the other v is the unit vector indicating the direction of the velocity (the negative sign indicating the drag is opposite to that of velocity).
The reference area is the area that is the "face" of the object as it falls. Think of the front of your car. If your object is a block, it will orient itself to fall with the smallest face forward. So 1/2" by 1/2" or .25" square will be the area.
You calculate the weight of the object underwater. Substitute the weight of the object for Fd (weight of the object will equal drag).
Then you solve for v. That is the velocity you are looking for.