Question

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.)

Explanation:

Step1: Analyze Initial State (A)

At \( t = 0 \) seconds, the skydiver just starts falling. Air resistance (\( F_{air} \)) is 0 (since velocity is 0, and air resistance depends on velocity for a skydiver, \( F_{air} \propto v^2 \) or \( F_{air} \propto v \) depending on model, but initially \( v = 0 \)). The only force is gravity (\( F_g \) downward). Net force \( F_{net}=F_g - F_{air}=F_g - 0 = F_g \), so net force magnitude is equal to weight.

Step2: Analyze 15 Seconds (B)

After 15 seconds, the skydiver is accelerating downward, so velocity \( v>0 \), so \( F_{air}>0 \), but \( F_{air}0 \)). So net force magnitude is \( F_g - F_{air} \), which is less than \( F_g \) (from step 1), but still positive.

Step3: Analyze 45 Seconds (C)

As time increases, velocity increases, so \( F_{air} \) increases. At 45 seconds, she is approaching terminal velocity (reached at 60s). So \( F_{air} \) is closer to \( F_g \) than at 15s. Thus, net force \( F_{net}=F_g - F_{air} \) is smaller than at 15s, but still positive (since not at terminal velocity yet).

Step4: Analyze 60 Seconds (D)

At terminal velocity, the skydiver's velocity is constant, so acceleration \( a = 0 \). By Newton's second law, \( F_{net}=ma = 0 \). So net force magnitude is 0.

Summary of Net Force Magnitudes:

• A: \( F_{net}=F_g \)
• B: \( 0 < F_{net} < F_g \) (and larger than C's net force)
• C: \( 0 < F_{net} < F_g \) (smaller than B's, larger than D's)
• D: \( F_{net}=0 \)

So the order of net force magnitude (from largest to smallest) is \( A > B > C > D \), and D has net force 0.

Answer:

To identify the net force vectors:

• A (t=0): Net force vector (downward) has magnitude equal to weight (gravity, since \( F_{air}=0 \)).
• B (15s): Net force vector (downward) has magnitude \( F_g - F_{air} \) (smaller than A, larger than C).
• C (45s): Net force vector (downward) has magnitude \( F_g - F_{air} \) (smaller than B, larger than D).
• D (60s): Net force vector magnitude is \( 0 \) (since \( F_{air}=F_g \), forces balance).

(If identifying relative magnitudes: \( A > B > C > D \), with D having net force \( 0 \).)

(Note: Since the vector diagrams are not shown, but based on the physics, the above analysis gives the relative net force magnitudes for each position.)