In reply to:
.........The horizontal distance from an object after a certain delay depends only from your "velocity at exit", and so it depends really so much on each jumper. But some time ago I made few calculations making a guess. I supposed the following.
Suppose to have a jumper that performs a "long jump from still" to his "maximum length possible at the given exit velocity" (I know in physics it has another specific word, but in this moment I am missing the English word for it), that is, the jumper jumps with a trajectory of 45° on horizontal plan, and let us suppose this jumper, with his jump to his "maximum length possible", reaches a jump of 2 m, that is, final position on horizontal plan of his feet is 2 m from position at exit. A "long jump from still" of 2 m is doable. Provided what above, let us have V has "velocity at exit" and Vh as "horizontal velocity at exit", and having √2 = 1.41 and g = 9.8 m/s², we have the following results:
V²
— = 2 m after t = 0.64 s · · · V = √ (2 · g) = 4.427 m/s
g
Vh = (√ 2 / 2) · V = √ g = 3.13 m/s
With the above results, we can now write the distance, x (t), "flown" horizontally after 0.5 s, 0.64 s, 1 s, 1.5 s, 2 s, 2.5 s, 3 s, 3.5 s, 4 s, 4.5 s, 5 s, etc. of freefall:
x (0.5) = 1.6 m
x (0.64) = 2 m
x (1) = 3.1 m
x (1.5) = 4.7 m
x (2) = 6.3 m
x (2.5) = 7.8 m
x (3) = 9.4 m
x (3.5) = 11.0 m
x (4) = 12.5 m
x (4.5) = 14.1 m
x (5) = 15.7 m
Practical example: if you exit with the above velocity and if your canopy opens 3 s after exit, you are 9.4 m far from the object.
Of course, provided the modest horizontal speeds involved, we can neglect any influence of jumper's body with respect to air drag; and it does not take into account at all any horizontal speed you could gain tracking, but that (tracking) would only help, still I do not think you can gain any significant tracking effect below 3 s of freefall (while if you start to track immediately after exit you can have a benefit in the sense the track will be as much efficient as possible, as soon as possible...).
If the jumper runs at exit instead of doing a "long jump from still", what you have to do is calculate (=guess?) your horizontal speed at exit and then do your homework getting the horizontal distance "flown" multiplying the horizontal speed times the elapsed time.
Another guess: if the "long jump from still" gives you a horizontal exit velocity of 3.13 m/s and the fastest men on earth can run at 11 m/s (top velocity), as a run of a BASE jump, with your trekking shoes, clothes, weigth of rig, PC handheld, limited distance for running, etc. etc., reasonably you could have a horizontal exit velocity of 4 m/s, 5 m/s, 6 m/s at best, in any case hardly 7 m/s or more than this...
For conversion meters/feet, provided that 1 ft = 0.3048 m/ft and 1 m = 3.2808 ft/m, you can help yourself.