A paratrooper steps out of an airplane at a height of 1000 ft and after 5 seconds opens her parachute. Her weight, with equipment, is 195 lbs. Let y(t)denote her height above the ground after t seconds. Assume that the air resistance is 0.005y′(t)^2 lbs in free fall and 0.6y′(t)^2 lbs with the chute open. At what height does the chute open? How long does it take to reach the ground? At what velocity does she hit the ground? (This model assumes that air resistance is proportional to the square of the velocity and that the parachute opens instantaneously.)

(Hint: This problem can be solved most efficiently by using an ODE file to detect significant events. Pay attention to units. Recall that the mass of the paratrooper is 195/32, measured in lb sec2/ft. Here, 32 is the acceleration due to gravity, measured in ft/sec2.)

(b) Let v = y′ be the velocity during the second phase of the fall (while the chute is open). One can view the equation of motion as an autonomous first order

ODE in the velocity:

v′ = -32+(192/1950)v^2

￼Make a qualitative analysis of this first order equation, finding in particular the critical or equilibrium velocity. This velocity is called the terminal velocity. How does the terminal velocity compare with the velocity at the time the chute opens and with the velocity at impact?

(c) Assume the paratrooper is safe if she strikes the ground at a velocity within 5% of the terminal velocity in (b). Except for the initial height, use the parameters in (a). What is the lowest height from which she may parachute safely? (Please do not try this at home!)

(Hint: This problem can be solved most efficiently by using an ODE file to detect significant events. Pay attention to units. Recall that the mass of the paratrooper is 195/32, measured in lb sec2/ft. Here, 32 is the acceleration due to gravity, measured in ft/sec2.)

(b) Let v = y′ be the velocity during the second phase of the fall (while the chute is open). One can view the equation of motion as an autonomous first order

ODE in the velocity:

v′ = -32+(192/1950)v^2

￼Make a qualitative analysis of this first order equation, finding in particular the critical or equilibrium velocity. This velocity is called the terminal velocity. How does the terminal velocity compare with the velocity at the time the chute opens and with the velocity at impact?

(c) Assume the paratrooper is safe if she strikes the ground at a velocity within 5% of the terminal velocity in (b). Except for the initial height, use the parameters in (a). What is the lowest height from which she may parachute safely? (Please do not try this at home!)

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