If the MH370 black box did sink to 15000 ft, how long would it have taken? I have been following MH370 on the news just as everyone and latest reports seem to indicate that the black-box could be found. A recent info-graphic http://t.co/lyBBE9C2hF shows the insurmountable depth of the oceans and how the black-box could have sunk 15,000 ft! I wonder how long it would have taken for it to sunk to the bottom of the sea-bed? What is the equation of motion for a sinking object at sea, ignoring under-water currents? 
 A: The relevant equation is the kinematics with linear drag. In this case, there is a resistant force that acts opposite gravity (i.e., upwards) and is linear to the velocity at which it travels:
$$
\mathbf F_D=-b\mathbf v
$$
where $b$ is some fluid- and object-dependent constant.
Using Newton's 2nd law,
$$
m\ddot{\mathbf x}=m\mathbf g - b\dot{\mathbf x}
$$
If we assume a one dimensional case,
$$x(t) = \frac{c_1 m e^{-b t/m}}{b}+\frac{g m t}{b}+c_2$$
If you know what the constants are (depends on assumptions at the boundaries, e.g. was it stationary or moving), then you can figure out the time it took. 
A: You say, ignore currents and I assume, other extraneous factors.
If that is the case, then, considering a hydrostatic balance in the water column where z is the vertical coordinate. Then, the motion of a water parcel with density, $\rho$, displaced upwards by a distance, $\Delta z$, in a fluid with a reference density, $\rho_0$, is governed by
$\rho_0 \frac{d^2\Delta z}{dt^2} = g \frac{d \bar{\rho}}{dt^2} \Delta z$
I am confident you will be able to deduce the rest.
