Are you familiar with the equations relating acceleration to velocity and displacement?

\(\displaystyle V=U+At\)

\(\displaystyle D=O+Ut+1/2At^2\)

Where:

A is acceleration

t is the elapsed time

U is initial velocity (when t=0)

O is initial position (when t=0)

V is the velocity at time t

D is the displacement at time t

The acceleration is given by F=mA (or A=F/m)

Where F is the Drag created by the parachute m is the mass of the package.

That is horizontal velocity and displacement,

You also have a second set of these equations for vertical displacement, but here the acceleration is due to gravity,

Note that as the package falls you might need to consider how much drag is acting in the vertical direction and how much in the horizontal direction.

This will change as the angle of the trajectory of the package changes.

Watch out for your reference point, is it on the ground or in the aircraft?

Either, or even both may be used, (but be careful if you try using both, just use one at a time)

(It is even possible to use a reference point based in the package, but non-inertial reference points can get confusing).