Hmmm,

The way I would approach the problem would be to calculate the

terminal velocity in water, assuming you reach the terminal velocity fairly rapidly.

(terminal velocity equation without buoyancy)

[attachment=18400]

V

_{t} is the terminal velocity

m is the mass of the falling object

g is the acceleration due to gravity, 9.8 m/s

^{2}C

_{d} is the drag coefficient

ρ is the density of the fluid (1.027 gm/cm

^{3} for seawater)

A is the projected area of the object.

Ok, let's start with a shot put.

m = 7.260 kg

C

_{d} = 0.47 (sphere)

diameter: 120mm (0.12m) (area: 0.0113 m

^{2})

So,

V

_{t} = sqrt((2*7.260 kg*9.8 m/s

^{2})/(1027 kg/m

^{3} * 0.0113 m

^{2} * 0.47 ))

V

_{t} = 5.1 m/s

The maximum depth of the Mariana Trench is: 10.911 km (10,911m)

(10911/5.1)/60 = 35.7

**So, it takes your shotput about 36 minutes to fall to the bottom of the Mariana Trench.**Let me try the equation for terminal velocity with buoyancy (for a sphere) at the bottom of the Wikipedia page above.

'

(terminal velocity equation for spherical object with buoyancy)

[attachment=18401]

d is the diameter of the object (0.12m)

V

_{t} is the terminal velocity

g is the acceleration due to gravity, 9.8 m/s

^{2}C

_{d} is the drag coefficient 0.47

ρ is the density of the fluid (1.027 gm/cm

^{3} for seawater)

ρ

_{s} is the density of the object

A is the projected area of the object (sphere).

Calculating the density of the shotput, mass/volume using the measurements above.

(7.260 kg) / (4/3 π (0.06 m)

^{3})

8024 kg / m

^{3}V

_{t} = sqrt ((4 * 9.8 m/s

^{2} * 0.12 m * (8024 kg / m

^{3} - 1027 kg/m

^{3})) / (3 * 0.47 * 1027 kg/m

^{3}))

V

_{t} = 4.8 m/s

**And I come up with about 38 minutes to fall to the bottom of the Mariana Trench when including buoyancy**You can calculate the terminal velocity for different objects.

The calculation is dependent on the mass (based on the cube of the radius), and the frontal surface area (based on the square of the radius). So, a more massive object would tend to reach a higher terminal velocity, and fall to the bottom quicker.