The question needs some repairs before we can start answering it.
-- The answer depends on knowing how fast she was falling when the acceleration began. I'll assume the airplane was in straight and level flight when she just lazily tumbled out of it and began falling, so her initial vertical speed was zero.
-- Skydivers know how to stretch out their body or scrunch it up, to make themselves fall slower or faster. When they do that, they're USING air resistance. BUT ... since air resistance has an effect on their speed AND they can even influence and manipulate it, the only way we can calculate the distance of falling for 2 whole minutes is to assume that there's NO air resistance at all, and that's what I'm doing.
-- The formula appears incorrectly in the question. It should be
D = (1/2) · A · T²
Distance of fall from rest = (1/2) · (acceleration) · (time)² .
Swati's downward acceleration is the acceleration of gravity, on whatever planet she happens to be skydiving that day. The question says she's accelerating downward at 9.8 m/s². This is a dead giveaway ... it's all taking place on Earth !
So . . . . . D = (1/2) · A · T²
D = (1/2) (9.8 m/sec²) (120 sec)²
D = (4.9 m/sec²) (14,400 sec²)
D = 70,560 meters
There you have it ... As close as I'm able to tell, by doing the best I can with the information in the question; Swati free-fell for 2 minutes, and fell through a distance of 70.56 kilometers ... about 43.8 miles straight down !
The math is bullet-proof, but the situation is hockey pucky. An airplane can't fly at that altitude, and if it could, Swati would be foolish to jump out of it at that altitude. But, as I said, the math is bullet-proof.
