Question

A parachutist whose mass is 60 kg drops from a helicopter hovering 1000 m above the ground and falls toward the ground under the influence of gravity. Assume that the force due to air resistance is proportional to the velocity of the parachutist, with the proportionality constant b 30 N-sec/m when the chute is closed and b2 100 N-sec/m when the chute is open. If the chute does not open until the velocity of the parachutist reaches 15 m/sec, after how many seconds will the parachutist reach the ground? Assume that the acceleration due to gravity is 9.81 m/sec.

Answer 1

and Law of motion. By newtons deg hvn Law - Sht mtom log ( 7 v)+" at too, v=o; Arm 1099 os today 990] 0. When V= 15mls 9.81 Thus Distance ve on a ti = 2.89 seconds after 2.89 seconds, para chelte will open up. of covered - to = m lugl log(94) my lie tm) va

plan A (1-én) at LORM 4 ks] - @@ at t = 2.89 seconds, 7 x = 26-3 m ) A thus it has to cover now, 1000-26.212 973.29 m. B And now parcechulte is open. New K=100Ns/m Put these value in equation , r oot/60 973.29 = 60x9.81 e 1 + 100t] 1002 60 St + 3/3 - 226.594 atz ~ 274 seconds Total time of descent textz = 277 seconds. This

Please feel free to ask doubts in the comment section and give it a thumbs up if you liked the answer. Have a good day