Missing a few pieces to the puzzle.?

Question

Over the weekend I went on a saltwater fishing trip,and had a thought or two,but couldn't figure it all out.
Here's the thought: If I could get the fishing line with the heavy sinker to stop just before it hit bottom,I wouldn't have to wait for all the silt that washed up from it hitting bottom to clear up.How long would it take for a 10oz sinker to free-fall through 250feet of water(assuming no drag).
The physics formula for calculating air drop would be ...
D = ((1/2 AT^)V1T))
where D is distance
A is acceleration
T is time
V is Velocity
I remember A being 10meters/sec.,but that is for air test as it is acceleration to the ground.
250ft is just under 77 meters

some more insight:
Sky; we share the same line of thought.I thought since it was a gravity drop,you can eliminate the V1T part of the formula and work with D=1/2AT^ . Acceleration in this case is peak acceleration.It's obviousthat in the 1st 10-20 meters of descent that the sinker hasn't yet acheived that velocity.Also was there any "drag" created by the buoyancy of salt water vs air?The sinker was bomb shaped,and made an impressive 2-3 ft long tunnel of air bubbles when first dropped.Counting times by various ppl,it took between 18-25 sec for it to hit bottom at 250ft.Mind you this is with drag on the line by the fisherman.I just thought there was a little extra drag created by the buoyancy of salt water over air.(all else being equal)

Answer 1

The acceleration of gravity is still 9.81 meters/sec^2 even underwater but the "specific gravity" within a substance requires some knowledge of the objects volume. This is because of Archimedes Principle. Basically, the 10 oz weight is still being accelerated by nearly 10 m/s^2 gravity, but the amount of water it displaces may make the surrounding water push back enough to slow its apparent acceleration. If it displaces more than 10 oz of water, then it will not even sink but will float.

Without knowing the volume or specific gravity of the sinker, we cannot numerically answer your question. If an assumption was to be made, it would be that the sinker is likely a solid lead object whose volume would be small enough that one could estimate a minimal buoyant effect. Neglecting the drag of the water, and assuming an initial velocity of 0, the answer is:

D = Vi t + 1/2 a t^2
D = 1/2 a t^2 since Vi = 0
77 m = 1/2 10 m/s/s t^2
t = 3.9 seconds

In reality, it will take longer because it will displace some amount of water which will make the effective acceleration less, and it will experience drag from the water which will tend to slow the descent as well.

Expanding my answer with the new info:
The sinker will be slower falling through the water for three reasons that I can think of:
1. It will definitely experience more drag from moving through water versus moving through air. But the original question wording said "assuming no drag from water".
2. Archimedes Principle will cause its apparent acceleration to be less. Gravity is pulling down on all the water around the sinker and that water is also pushing against the sinker acting to buoy it up. I assumed the volume of the sinker was small compared to its mass so my answer would be the air time descent. It should fall slower because it will displace some quantity of water which will act to buoy it. This is not drag. Drag will act to oppose movement, Archimede's Principle will push it even if it is not moving and as I stated above, if its volume is large enough it will float and not even sink (think of a steel ship).
3. You mention it having air bubbles come off of it. If the sinker has cracks for the fishing line, or a cup shape to it, it may also trap air. This will also act to slow it since any air clinging to it will lower its density and thus its specific gravity. This again is Archimedes Principle.

Answer 2

ALWAYS