What is the formula for a parachute falling through the air?

Answer 1

Originally, it would follow the general kinematics formula of:
v = u + at
where u-initial velocity, v-final velocity, a-acceleration and t-time.
and
mg - R = ma
where m-mass of parachute, R-air resistance, a-resultant acceleration.

However, when R increases, there will be a stage when
mg - R = 0 => mg = R

At this instant, the velocity remains constant as there will not be anymore resultant acceleration. This is called the terminal velocity.

So the velocity, v is equal to the final velocity just before the terminal velocity is reached.

Answer 2

Interesting question: This is the best answer that I could whip up from the web (see sources)

Basically the drag of a parachute in the typical regime of Reynolds number is expressed as:

Drag Force = C * A * density * v^2 (opposed to the direction of velocity).

where C is an empirical coefficient (around 0.8 according to Dr. Patvin and others), A is the surface area, density is the mass/volume of the fluid (in this case air), and the v is the relative speed of the object in the fluid (parachute through the air). ..... [1]

Answer 3

R = mg
where R is the air resistant
m = the whole mass of the falling parachute
g = gravitational acceleration
assuming that the parachute falling with a constant velocity

Answer 4

Without a parachute it's s = u x t + 1/2 x g x t squared.

With a parachute it's s = u x t