# Describe and explain the vertical motion of a parachutist which jumps out of an aeroplane at time t=0 and then releases the parachute shortly after reaching terminal velocity at time t=T. (Assume air resistance is not negligible).

We need to understand both what this question is asking of us, and what the question tells us. We are asked to investigate the motion of the parachutist, therefore in this case we need to apply Newton’s Laws of Motion. Specifically, Newton’s second law F=ma. As we can describe the motion of the parachutist using their acceleration (a) it is clear that we have to investigate the forces the parachutist is subjected to. Let us define upwards as the positive direction. Therefore, as the gravitational field g is constant we know that the parachutist is subject to a constant gravitational force F(w) = -mg. We also know that as the parachutist falls with velocity v they exert a force on the air. Therefore, applying N3 we know that the parachutist must also be subject to an equal and opposite reaction force acting upwards. For air, this upwards force is proportional to both the square of the velocity and the effective surface area of the parachutist in contact with the air. Therefore, F(r) ∝ v^2 and F(r) ∝ A. Given all of this information we can actually answer the question now. At all times the resultant force F acting on the parachutist F = F(w) + F(r) = F(r) – mg. Therefore a=F(r)/m – g. At time t=0 the vertical velocity v of the parachutist v=0. Therefore F(r)=0 and so a=-g. Therefore, as the parachutist accelerates their speed begins to increase and thus F(r) also begins to increase. Therefore, the magnitude of the acceleration decreases until F(r)=mg and thus a=g – g=0. At this point the parachutist has reached terminal velocity at time t=T. The parachutist then releases their parachute and their effective surface area A increases drastically. Therefore, the resultant acceleration acts upwards and the speed of the parachutist begins to decrease until F(r) is again equal to mg and thus a=g – g=0. At this point the parachutist has reached a second, lesser, terminal velocity and continues to fall at this velocity until they reach the ground. My tutoring style is generally more discussion based and I would break this question down to make certain points clearer. Thanks!

Answered by Tutor43724 D. Physics tutor

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