For reasons given, the stone falls to the East. But, the estimates have ranged from 6 feet to 6 meters, so I tried to pin it down little better. In the solutions posted, the 6 ft calculation used a 1.6 km drop altitude and the 6 m calculation considered a drop from a balloon at 2 km. However, most nations generally limit everyday manned balloons flights to altitudes bellow 1 km. Also, air resistance, going as (v^2 A), will be a factor all the way down. E.g, if dropped from 1 km, the stone will just achieve its terminal velocity of 120 mph. The math, simple physics including air resistance, can be solved in closed form, but happily, I found an app that did it for me, and it gives the derivations too. 

For inputs to the app, I used a hardball-sized stone (r=2.7 cm, rho=2.7 gm/cm^3) which is dropped in air of density 1.16 kg/m^3.

(The range of possible densities, 1.11 - 1.26 kg/m^3, makes less than a 1/2 sec total time difference. 

For the stone V relative to the ground, I used: 

V = V(lat0) [(6371+alt)/6371 - 1] cos(lat), where V(lat0) = 460m/s. 

Below are listed the landing offsets for a range of altitudes and latitudes:

      Stone Displacement (w/air) (m)  descent time (s)   V(m/s)

                                       

      lat:    90      70      30   0(deg)  (w/air) (w/o air)* (wrt Earth)     

 alt  ---------------------------------      -----------------   -------

 300(m)    0.0m  0.07   0.17  0.19    9.0s   7.8s       0.022m/s

 500(m)    0.0   0.16   0.40  0.46    12.7    10.1        0.036  

1000(m)    0.0   0.53   1.34  1.55    21.5    14.3       0.072

1600(m)**  0.0   1.26   3.20  3.70    32.0    27.9      0.116   

 

2000(m)**  0.0   1.93   4.88  5.63    39.0    34.9       0.144

----------------------------------------------------------------------

notes: *(not used), **(unlikely)

From this table, a 1 km high stone would land displaced by 4.6 feet. 

Comparing this table with the previous (rougher) estimates, if the balloon had indeed been a mile high, air resistance would have caused a longer descent (32 s vs 18 s) and a displacement of 12 ft, as shown in the table.

Had the balloon been dropped from 2 km, the long descent would land it 5.6 m displaced. This agrees with the 6 meter value posted, but the agreement was in part due to a fortunate canceling of an underestimate of descent time (30 vs 39 s) and an overestimate of relative velocities (0.20 vs 0.14 m/s).

Nonetheless - the previous estimates were pretty good!