Physics, goofey question
 

How fast will an object (3-4 oz.) fall in water? Anyone know?

Thats in Ft./ second

It depends on the mass of that object. A solid lead ball will fall a lot faster than a hollow lead sphere. Obviously a 4 oz chunk of balsa wood will not fall at all.

If the object's density is less than 62.4lbs/cubic foot, it won't fall at all - it will float! ;)

You need to specify much more information to get a decent answer and even then it's going to depend on the shape of the object, and a host of other things.

If it's your keys it's too late. You ain't gettin em back.

LMAO.....good one.

Yeah, this is even worse than the same question about something falling in air. Heck, I think the temp of the water could even have an affect (though probably negligible under normal circumstances). Too many variables. It's probably more dependent upon shape, density and the object's surface than it is on the objects weight/mass.

not enough info. shape of the object? ohhhh...steve has it.

She's a duck!

wood floats, ducks float. Witches are made of wood.

1" L x 1/2" W x 1/2" high, Fresh water, 4 ounce weight, in Ft./sec

Also dips into a little chemistry. Salinity and density of the water play a part.

African or European?

Your object has a volume of 3.98 cm3 for a weight of 113.39g, so density is 28.5 g/cm3.
Huh, that is very dense, denser than uranium (unless measurement were feet, not inches) Will definitely sink, but what could it be?

The doctor of applied physics who is looking over my shoulder right now (she's hot too!) says, "It will fall as fast as it can."

:D

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:

http://forums.pelicanparts.com/uploa...1263920939.jpg

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.

and it is accelerating as it falls, so it depends on where/ when it is measured, till it reaches terminal velocity in the h2o

wow , info overload. at work, Diamond drills- exploration mining we sometimes have to dry sharpen our bits, we use diamond bits hence the name, at depths over 2000 feet it is real tricky and the oppertunity to f up is huge, hyd. delay, rod stretch (2" for 1000 feet) etc... so I get the guys to drop a segment from an old bit down the hole and as it gets cut up it will strip the bit on the rod string.
What is it made of, Diamonds, synthetic diamonds, carbide and tungston dust, carbide inserts and yes they are heavy.
Reason I asked this question was for timing purposes, when we drop the segment how long would we wait before we dropped a piece of our drilling equipment that is regulated to fall at 100 ft./min, will the segment fall faster than 100ft. per minute?

Newton's apple and the feather fall accelerate towards the earth at the same rate, as i recall.

It's even more complicated then, if the object is falling down a narrow hole and has to displace water which has a limited place to go? Think oil damped shock absorber.

Not to be a smarm, but I think you might turn to experimental physics rather than theoretical. Find a deep enough lake, attach (firmly) a tether of marked and measured length to the old bit, and see how fast it falls, time at 20 feet intervals, figure out when it stops accelerating and the velocity after that point. Then drop it through a steel pipe placed vertically in the lake, measure the difference in velocity with and without pipe, even if it isn't at terminal velocity before reaching the end of the pipe, you'll still get an estimate of how much the pipe slows it down. Sounds dumb but that's what I'd do, even if you have a calculated theoretical result it would be best to verify experimentally.