What are all the factors you would need to calculate the terminal velocity of a parachute?

Question

Attached to a person, of course.

Answer 1

I will need
- the parachute,
- the person, and
- a cliff high enough.

The factors will be distance ∆h travelled in time ∆t.

Answer 2

I assume you define terminal speed as the speed with which a falling object impacts the ground. And that it's about a simple, unsteered, lift-free chute. You can approximate the terminal speed as the equilibrium speed which can be determined analytically from the air density and grav. acceleration at ground level, and drag coefficient, mass and area of the loaded chute. But determining the true terminal speed would take a high-fidelity simulation and addionally knowledge of the variation of air density and grav. acceleration with altitude, and the release height.
You may wonder why the last item is needed. As altitude decreases the air density increases faster than gravity does so equilibrium speed decreases. During initial acceleration in thin atmosphere, the chute approaches and then exceeds equilibrium speed. Somewhere here it reaches the maximum speed of the fall. For the rest of the fall it decelerates but never converges exactly on the decreasing equilibrium speed. The drag force must exceed the grav. force and the difference is what slows the fall. Initial conditions affect the closeness of a converging function; however, the effect washes out over time and has to be infinitesimal over a few minutes descent. (Even if the air were constant density and g were constant, the approach to the constant equilibrium speed would be asymptotic, so you can see that initial conditions would still play a part.)