# How can I solve this Mechanics problem, about a cylinder in water?

1. A cylindrical submarine has a radius R of 3.2 m, a length L of 35 m and a total mass of 1.2×10^6
kg, of which 1×10^5 kg is ballast that can be jettisoned. The total upwards buoyancy force, B (N), is regulated as a function of depth and is given by
B =(1020 + 0.25z)Vg ,
where z is the depth from the water surface to the top of the submarine (m), V is the volume of the submarine (m^3) and g is the acceleration due to gravity (9.8 m/s^2).
(i) Neglecting viscous drag find the maximum downward velocity of the submarine assuming it
descends from a stationary position just below the water surface. The volume remains constant.
(ii) If the captain decides to drop the ballast at a depth z = 250 m, how much deeper (than 250 m) will the submarine sink?

this is all assuming the submarine is shaped like a simple cylinder. Can you help me?

2. Homework Equations the one given and F=ma is all that i can think of

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well, we need to know the volume of the submarine. so the equation for the volume of a cylinder is helpful....

probably: pi*r^2*L...

I'm not sure, but does the equation for B give us an acceleration? I don't think anything cancels out the /s^2 part of gravity so i think we get an acceleration for B.

isn't there an equation something like V = 1/2acceleration^2 + V(0).... If we plug our result for B into this... I think there's a (t) in the equation above. don't know exactly where this is going.

well... just some thoughts....

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