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Question

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A parachutist whose mass is 100 kg drops from a helicopter hovering 3000 m above the ground and falls under the influence of gravity. Assume that the force due to air resistance is proportional to the velocity of the parachutist, with proportionality constant b3 = 20 N-sec/m when the chute is closed and b4 = 100 N-sec/m when the chute is open. If the chute does not open until 30 sec after the parachutist leaves the helicopter, after how many seconds will he hit the ground? If the chute does not open until 1 min after he leaves the helicopter, after how many seconds will he hit the ground?

Explanation / Answer

mg-bv=ma

==> mg-b*v=mdv/dt

==> mdv/(mg-bv)=dt

now integrating from 0 to v

==> -m/b *ln[(mg-bv)/mg]=t

==>(mg-bv)=exp(-bt/m)

==> v=[mg-exp(-bt/m)]/b

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