a skydiver is dropped out of an airplane at an altitude of 10000 feet. she reaches a terminal velocity 60 seconds later. consider four positions during her fall. a: initial state (t = 0 seconds) b: 15 seconds after drop c: 45 seconds after drop d: 60 seconds after drop toggle through the set of vector diagrams at the right to identify the relative magnitude of the net force vector for each of these four positions. (consider vertical motion only.)

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Accepted Answer

# Explanation: ## Step1: Analyze Initial State (A: t=0) At $ t = 0 $, the skydiver has just been dropped. The only vertical force is gravity ($ F_g $) downward, and air resistance ($ F_{air} $) is 0 (since velocity is 0 initially). So net force $ F_{net}=F_g - F_{air}=F_g $ (downward), magnitude equal to $ F_g $. ## Step2: Analyze 15 Seconds (B: t=15) After 15 seconds, the skydiver is accelerating downward, so velocity is increasing. Air resistance is proportional to velocity (or velocity squared, but for relative magnitude, we know $ F_{air}0 $). So $ F_{net} $ magnitude is less than at $ t = 0 $, but still positive (downward). ## Step3: Analyze 45 Seconds (C: t=45) As time approaches 60 seconds (terminal velocity), velocity is closer to terminal velocity, so air resistance is closer to $ F_g $. But since terminal velocity is reached at $ t = 60 $, at $ t = 45 $, $ F_{air}F_{air1} $) - D: $ F_{net}=0 $ (smallest) So the order of net force magnitude (from largest to smallest) is $ A > B > C > D $, and at D, net force is 0. # Answer: - A: Net force magnitude = $ F_g $ (downward) - B: Net force magnitude $ < F_g $, downward - C: Net force magnitude $ < B $, downward - D: Net force magnitude = 0

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QUESTION IMAGE

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Explanation:

Step1: Analyze Initial State (A: t=0)

Step2: Analyze 15 Seconds (B: t=15)

Step3: Analyze 45 Seconds (C: t=45)

Step4: Analyze 60 Seconds (D: t=60)

Relative Magnitudes:

Answer: