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Bowling Ball Release

How would I calculate the time it takes for a 12 lb bowling ball to sink to a 5 mile depth in the ocean?
If I was on a boat at a location in the ocean that was 5 miles deep, and I released 12 lb bowling ball from the surface, how do I calculate the time it takes for it to reach the bottom?
I'm guessing that the variables include the viscosity of the water and the constant of gravity and the surface of the ball itself.
And before anyone comments on it , this actually isn't a homework question, despite the fact that it sounds like one. This question is simply out of personal curiosity.
Well, an object's buoyancy is equal to the weight of the fluid it displaces, and acts with a force equal to that weight. So, a bowling ball will sink with normal gravitational acceleration (9.8 m/s^2) minus the force pushing it up by the water it is displacing, until it reaches its terminal velocity, when it will to continue to sink at that rate until it hits the bottom. Kind of.
The density of sea water is highly variable, and taking pressure, temperature, and salinity into account makes a calculation like this very tedious. The best way to go about it is to assume an average density for seawater for the whole trip down. The calculation will be simpler, but the answer should be moderately accurate. Water at the surface has a density of about 1.025 g/mL, and deep, cold water about 1.05 g/mL, so 1.0375 g/mL is an acceptable value to use.
The best way to see how an object sinks is to calculate its net force. This is just the weight of the object (downward force) minus the buoyancy of the object:
Fnet= mg - pVg,
where m is the object's mass (5.44 kg for a 12 lb ball), p is the fluid density (1037.5 kg/m/s^s), V is the object's volume (4/3pi times it's radius [.109m for a standard ball] cubed), and g is the coefficient of gravitational acceleration (9.8m/s^2 on Earth).
So, Fnet = (5.44kg)(9.8m/s^2) - (1037.5kg/m^3)(4/3pi[.109]^3)(9.8m/s^2)
The answer, in a unit called Newtons ([kg m]/s^2), is 47.6839. But what does it mean?
The net force on the ball is 47.6839 Newtons toward the bottom (force is a vector, and therefore has directionality, but more simply, we know the bowling ball is sinking). Recalling another of Newton's equations will allow us to calculate the acceleration of the ball. Specifically, F=ma.
We already know the Force and the mass, so we can plug those in and solve for acceleration. To make things simpler, we'll move the formula around before we put the numbers in it.
F=ma, divide both sides by m,
and F/m= a. So if we divide the Force by the mass, we'll get the acceleration. This makes sense, remember our units of Force are newtons, kg m/s^2, so if you remove the kg, you just have m/s^2, the basic unit of acceleration.
So, 47.6839N/5.44kg = 8.765m/s^2 This means that every second, the ball will be moving 8.765 m/s faster than it was the second before.
Now, all we have to do is find out how long the ball will accelerate before reaching terminal velocity, and just put everything together.
The formula for calculating terminal velocity is
Vt= (sqrt[2ma/pAD])
where m is the objects mass, a is the object's acceleration, p is the fluid density, A is the surface area of the object (4pi[radius^2], and D is the drag coefficient (.47 for a sphere).
so, for our situation,
Vt = (sqrt[(2)(5.44kg)(8.765m/s^2) / (1037.5kg/m^3)(4pi.109m)(.47)])
Vt= .378m/s
This means that the fastest the ball can sink is .378 meters per second, which seems about right. That's about a foot and 2.5 inches a second.
This low terminal velocity is convenient, since we know the bowling ball (because of its 8.765m/s^2) will reach it well in under a second, we can just assume this speed for the entire trip.
Now, all we have to do is see how long the ball takes to travel 5 miles (8,046.72 meters) at this speed.
8,046.72m / .378m/s = 21,287 seconds = 354.79 minutes = 5 hours, 54 minutes, and 36 seconds
So, a 12 pound (5.44 kg) bowling ball, dropped into 5 miles of ocean (8,046.72 meters) will accelerate until it reaches a velocity .378m/s, at sink straight to the bottom in 5 hours, 54 minutes, and 36 seconds.
Roy D Mercer - Bowling Ball Fungus
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