If you were on a sailboat above the deepest part of the ocean, how long would it take a 50lbs cannon ball?

Question

... to reach the ocean floor. I was asked this question in an interview. They were testing my thought process and I am very interested in a rough answer!

Answer 1

what a great question....
will it fall at the same rate when the water gets so dense? at 10,000 meters?

Answer 2

since this was an interview question, I'll assume that an answer that doesn't use reference material is actually a better answer

deepest part of ocean, don't know so lets guess that the trenches below sea level aren't much bigger than the mountains above seal level and lets say, 5 miles (about 25000 ft)

objects should reach terminal velocity in water very quickly so the ball will fall at about the same speed the whole way

I have dropped heavy things in water before so I should be able to visualize about how long a cannon ball would take to hit the bottom of the pool deep end (about 10 feet)

It seems like it might take about 1 second
so maybe that is a good range

about 2500 seconds to the ocean bottom

for easy math, call it 2400 seconds for the short end
that would be, hm, 40 minutes

I'd say between 1/2 an hour and an hour

Answer 3

What's the radius of the cannon ball?
What's the weather like? (sea temperature)

I guess, the terminal velocity calculation will have to suffice. But a the one from Stoke's!

v = (2 r**2 g (d1-d2) )/ 9c

v is the terminal velocity
r is the radius of the sphere
d1 is the density of the sphere (probably about 8)
d2 the *density* of seawater (about 1)
g is gravitational acceleration
c is viscosity of the fluid in (in Poise?)

But the pressure!!!
Well pressure changes enourmously but not the density for water(not very much at least) - so the water's viscosity can be treated pretty uniformly (except for temperature variations) - water at 100psi or 10,000psi is practically the same viscosity wise.. Weird but true, amaze your friends. Also weird but true is some other fluids, like oils, compress with pressure and so viscosity increases like you'd might expect.

So is that all?
Maybe not - if this all takes place at speeds which disturb the laminar flow around the cannon ball (is it just a sphere?) then we're ok..(no, don't ask me to calc that) But if laminar flow is disrupted you have all that crazy turbulence and drag goes through the roof Reynolds comes and lays some smack down... erm... I mean you can calculate Reynolds numbers (Osborne Reynolds...supppah genius) - if they're small then you're safe with Stoke's .. if not call NASA.

That it? Umm, viscosity can be changed by temperature, of course, so you know this basically comes down to the fact that is a ridiculous question to ask for an answer but a fun one to ask if you just want to see someone's thought process.

Answer 4

The terminal velocity of a cannonball in water would be about 12 m/s, the Challenger Deep is 11000 m deep, so it would take about 15 minutes.

Answer 5

It depends on how much powder they put in the cannon when they fire the thing.