A person steps out of an airplane at a height of 1000 ft and opens her parachute after 5 seconds. Her weight, with equipment, is 195 lbs. Let [tex]y(t)[/tex] denote her height above the ground after [tex]t[/tex] seconds. Assume that the force due to air resistance is [tex]0.005y(t)^2[/tex] lbs in free fall and [tex]0.6y(t)^2[/tex] lbs with the chute open.

1. At what height does the chute open?
2. How long does it take to reach the ground?
3. At what velocity does she hit the ground?

(Note: This model assumes that air resistance is proportional to the square of the velocity and that the parachute opens instantaneously.)

The physics of a parachuter's fall includes gravity and air resistance, which changes when the parachute is deployed five seconds into the free fall. To calculate the specific information needed, we have to consider her initial velocity, the time-integrated acceleration due to both air resistance and gravity, and the equation modeling her height over time. The calculations involve numerous physics concepts, such as force balance and acceleration calculation.

Explanation:

The question involves the physics of free fall and air resistance. When the parachutist jumps out of the plane, she experiences gravitational force. This force is represented by her weight which is 195 lbs. For the first 5 seconds of her fall, air resistance is modeled as 0.005y(t)² lbs, where y(t) is her height at time t.

We know that she opens her parachute after 5 seconds, so y(5) will be the height at which the parachute opens. The situation changes after the parachute opens, as the air resistance increases to 0.6y(t)² lbs.

To calculate the height when the parachute opens, we would need to know her initial velocity and integrate the acceleration (caused by gravity and air resistance) over the given time.

Similarly, to know the time it takes for her to reach the ground, we would need to solve for t when y(t) is 0, considering both periods of her fall (free fall and after the parachute opens). Lastly, the velocity at which she hits the ground can be calculated from her final acceleration at the point when y(t) equals 0.

During her fall, two main forces are in effect: gravity pulling her down and air resistance opposing this motion. The balance of these forces changes dramatically when the parachute is deployed, causing her to decelerate.

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