How long does it take to reach maximum freefall speed?

Question

If you jump from an airplane, how many seconds/what drop distance before you are travelling at maximum speed.

I understand the maximum freefall speed is around 120mph.

Answer 1

The maximum speed is about 220-235 MPH, and it can be achieved in about 20 seconds of freefall, in a verticle dive (hands flat against the legs, body very straight, head down attitude), which would require only about 6000 feet of altitude. Then the parachutist would have to assume the 'flat and stable' body position, after about 25 seconds, in which the back is arched, arms outspread, hands flat, facing the earth, legs slightly spread and slightly bent. This will allow him/her to slow down to the normal 120 MPH speed for the opening of the parachute at a safe altitude. To open a parachute at the higher speeds could easily cause sturctural failure of both the parachute, and the human body. In civilian skydiving, 2500 feet is considered the minimum safe altitude for parachute deployment. Only licenced Experts are allowed to go lower, in the US the minimum for the "D" (Expert) licencee is 2000 feet. In Europe it is 600 meters, or 1960 feet. Many military drops are conducted at lower altitudes, but rarely with freefall rigs. The openings on these parachutes are automatic and happen immediately as the jumper leaves the aircraft. Federal Aviation Regulations do not regulate military operations.

Answer 2

Lot many issues are not stated
in the question.The altitude of the
airplane is not specified. If it is a
transport plane it flies around
35000ft. If it is a fighter plane it
could be as low as 5000ft. But it
has escape systems. The equations
governing free fall is as under.
s=u*t+1/2*a*t^2
v^2-u^2=2as
u[vertical]=0
v=final velocity
a=32ft/sec^2
It must be remembered that for safe
opening of parachute[without tearing
] you can not have speeds above 120mph.
Hence the free fall has to be between
35000 to a level where the velocity
attained is 120mph or 176ft/sec.
v^2-u^2=2as u=0
176^2=2*32*s
s=176*176/64
=484 ft
time of fall
16t^2=s
t^2=484/16
=30.25
t=5.5sec

Answer 3

about 3 seconds before the drag from air friction equals the 9.8 m/s^2 acceleration from Earth gravity

Answer 4

You would fall for about eleven hundred feet to maximum free fall speed,