Whether the pipe bottom end is capped or not, the phenomenon depends only on the actual gap between the object and the pipe: if this gap is really small, then the flow around the object would be laminar and I think it is quite difficult to make an estimate under these conditions. Otherwise one could figure out a pressure drop coefficient for the flow in the restriction around the object and equate the pressure drop (that will somehow be proportional to the velocity squared) times the front area of the object to the driving force (weight minus Archimedes): this will give the constant drop velocity that will be presumably reached after a few feet of travel.
Is this a real life problem or a school problem?
If it is a school problem please keep away from here.
Otherwise you should give us some more figures (tube diameter, object dimensions and shape, gap around the object, expected falling velocity minimum and maximum value) to help figure out a method for the calculation.

prex
motori@xcalcsREMOVE.com
http://www.xcalcs.com
Online tools for structural design

Prex,
This is a real life problem, a crown plug was dropped from 100' through tubing onto one of our closed valves. I am trying to determine the resultant force the plug would invoke onto the valve to prove to a client that it would be sufficient enough to cause noticeable damage.
The ID of the tubing is 6.094" and we are taking the OD of the crown plug to be 5.90". The estimated weight of the plug is 40kg. We are taking the plug as a cylindrical shape, even though this is not the case, this would give us an approximate value which is what we are looking for.

M.

Try this as a route (I confess I'm cobbling it together quickly because I have to go so I've not done a longhand version to see how it works)
d = dia of plug
area of plug = A1 = pi *d^2 /4
D = dia of pipe
area of pipe = A2 = pi * D^2 /4
Water flow area past the plug = Aa = pi * [D^2 - d^2] /4
Wetted area = C = pi * [D + d]
hydraulic dia of annulus = Dh = 4 * A / C
B = flow area ratio = Aa/A2
length of the plug = L
rho = density of the water
to fall dL, the water displaced is dQ =[pi * (D^2 - d^2)] *dL
speed of fall = dL/dt
the displacing pressure is Pdisp = {40 kg / [pi * d^2]}
this equals the pressure loss across the plug
(initially I'm ignoring friction losses because they're too difficult to estimate quickly and probably are insignificant compared to the end losses)
The velocity of flow in the annulus = V = dQ /A
the velocity head in the annulus = hv = rho * V^2/2g (g = gravitational constant if you're using engineering units)
Consider that the inlet loss into the annulus is deltaP1 = 0.5 *(1-B)* hv
and the outlet loss from the annulus is deltaP2 = (1-B)^2 * hv
deltaP = deltaP1 + deltaP2 = Pdisp
equate these and produce the equation representing the speed of fall

Find the speed of fall at the bottom of the pipe = v
and just dL before the bottom of the pipe = u
Then from Mr Newton we get v^2 = u^2 + (2 * f * dL)
and the force is then Force = mass * f

Hope this points you in a good direction
Let me know if it works

This could be approximated by using a rotameter equation. Try this link:

http://egweb.mines.edu/eggn350/spring02/flow.htm

and go about halfway down the page to see if some parts of the rotameter equation can apply.

Flareman did a fine detailed development, however I'll propose a more simplistic approach.
Assuming turbulent flow, consider a head loss coefficient of 0.5 for the sudden contraction at object front and a coefficient of 1 for the sudden enlargement at object rear.
The pressure drop across the object is
deltap=1.5rho*v²/2
At the constant speed that will develop after the initial acceleration, the deltap will counterbalance the driving force (weight minus lift) divided by the object front area.
Now introducing the numbers and assuming your object is made out of steel:
Weight: 40x9.8=400 N
Lift: 50 N
Object area: 17640 mm²
Pressure drop: (400-50)/17640 = 0.02 MPa = 20000 Pa
rho=1000 kg/m³
v=sqrt(20000*2/1000/1.5)=5 m/s
This is the answer to your problem.
However this is valid under certain conditions:
1) The flow must be turbulent. To check this, one may calculate D_h=10 mm and, assuming 1 cp for the viscosity of water, and Re=50000. The flow is indeed turbulent.
2) The shape of the object must be purely cylindrical (not shaped as a missile), otherwise the enlargement and contraction losses will display coefficients lower than 1.5 as a total. This also explains why my solution does not depend on the pipe diameter: this solution holds as far as the area around the object is much smaller than the pipe section area (but not too small, otherwise a laminar flow will arise).

prex
motori@xcalcsREMOVE.com
http://www.xcalcs.com
Online tools for structural design

Don't forget the force of buoyancy on the plug.  The force of buoyancy and the force of drag opppose the force of gravity in a force balance.

Marty - I like all assumptions and engineering estimations by others above. That said, the crux of the problem depends on the minute distance between the edge of the float and the walls of the tube, the relative roughness of the tube, the shape of the edge of the float. It also depends on whether or not there is convection from a difference in temperature between the bottom and top of the 100' tube. This is loaded with unknowns. We always want simple answers. The simple answers are usually based on assumptions that are far from actual conditions. The shear force between the edge of the float and the walls of the tube are a principle factor and one to most consider for further analyses.