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A parachutist whose mass is 75kg drops from a helicopter hovering 2000 m above the ground...

A parachutist whose mass is 75kg drops from a helicopter hovering 2000 m above the ground and falls toward the ground under the influence of gravity. Assume that the force due to air resistance is proportionality constant b1=30 N-sec/m when the chute is closed and b2=90 N-sec/m when the chute open. If the chute does not open until the velocity of the parachutist reaches 20 m/sec, after how many seconds will she reach the ground use g=9.81

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Lety (z) be the parachutists height above ground so, y (0)-2000,y(o)-. Now by the Newtons second law: then. Assume v = by 77m The integratibg factor will be e Multiply this integrating factor both sides of V + by g 12 /n So, v--Ce This analysis is valid both for open and closed parachute. Now use v(0), to get for the first, rapid fall mg 575(10 75 so, v(t)25+25e5 Find 4 when v(4)--20 25, 1 10 5 after that time we get, v=--+Gem Use v(t)--20, to find Ca 402)5 C-1458.33 But,yv When parachute is closed: Use y (0) 2000 75(25) 30 )(4.02)-75(101402)_75(25),38 When parachute is open: + 2062.5 1949.48 30 30 Usey(,)-194948 (402)--75(10)(402)-75(25)e뿌+D, Finally, the parachutist reaches the ground when =1973.22 90 90 Db-2368s mg

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