The origin of your formula is the SUVAT equation for motion under a constant acceleration:

\(\displaystyle s = ut + \frac{1}{2}at^2\)

if u = 0 and a =g, then we can rearrange this for t:

\(\displaystyle s = \frac{1}{2}gt^2\)

\(\displaystyle t = \sqrt{\frac{2s}{g}}\)

However, this assumes a constant acceleration under gravity, which is not a suitable assumption for your parachute problem. You can't take a formula that makes that assumption and then retrofit an acceleration onto it("acceleration due to drag"). Instead, we have to lift the assumption that there is no air resistance and create a new formula.

I think you have two possible approaches to solve your problem. The first approach is the simplest, but also the most crude. The second is more complex, but will better model the whole system

- First approach: assume the parachute + payload just drop at constant velocity (terminal velocity). Then, \(\displaystyle s = vt\) will give you the solution, where v is the (constant) terminal velocity and t is duration of fall. It's a crude model, but it will have minimal errors in situations where the fall duration or distance is very large.
- Second approach: assume an equation for air resistance, develop an equation of motion based on Newton's laws and then solve the corresponding second order ODE.

The second of these approaches is mindful of the fact that air resistance is a force that contributes to a Newtonian dynamics problem, so in order to determine the fate of your parachute + payload, the drag force needs to be known. This drag force will compete with the weight of the parachute + payload and the result is the direct consequence of Newton's first and second laws. Most modellers doing this second approach assume an air resistance of the form \(\displaystyle F_r \propto v^b\) where b is usually something between 1 and 2. This, unfortunately, makes the maths rather difficult and sometimes non-linear.

Example of second approach: assume \(\displaystyle F_r = k v\), where k is some constant that calibrates the problem (case b=1).

Newton's second law:

Resultant force = mass \(\displaystyle \times\) acceleration:

\(\displaystyle Mg - kv = Ma\)

\(\displaystyle Mg - k \frac{dx}{dt} = M \frac{d^2 x}{dt^2}\)

\(\displaystyle \frac{d^2 x}{dt^2} + \frac{k}{M} \frac{dx}{dt} - g = 0\)

Now to solve this second order ODE. Because the constant term, g, this equation is imhomogeneous. The homogeneous equivalent is of the form:

\(\displaystyle \frac{d^2 x}{dt^2} + B\frac{dx}{dt}= 0\)

where \(\displaystyle B = \frac{k}{M}\). The auxiliary equation is consequently:

\(\displaystyle r^2 + Br = 0\)

\(\displaystyle r(r + B) = 0\)

solution are r=0 and r=-B. Since these are real and distinct, the trial solution is therefore:

\(\displaystyle x = c_1 + c_2 e^{-Bt} \)

However, this will only work for the homogeneous form. For the imhomogenous form, it turns out that the particular equation is \(\displaystyle \frac{g}{B}t\). Consequently, the full solution has the form:

\(\displaystyle x = c_1 + c_2 e^{-Bt} + \frac{g}{B} t\)

Let's confirm that this solution works. Differentiate to find derivatives:

\(\displaystyle \frac{dx}{dt} = - c_2 B e^{-Bt} + \frac{g}{B}\)

\(\displaystyle \frac{d^2x}{dt^2} = c_2 B^2 e^{-Bt}\)

Substitute these into our ODE to confirm that it is a solution:

\(\displaystyle c_2 B^2 e^{-Bt} - c_2 B^2 e^{-Bt} + g - g = 0\)

LHS = RHS, so it works. The full solution is therefore, substituting for B,

\(\displaystyle x = c_1 + c_2 e^{-\frac{k}{M}t} + \frac{Mg}{k} t\)

In order to eliminate the unknowns \(\displaystyle c_1\) and \(\displaystyle c_2\), you need to introduce boundary conditions (e.g. when t=0, v=0, x=0).

Note: x here is just s from the SUVAT equations.