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: Recall Terminal Velocity Concept

Terminal velocity is reached when net force (\(F_{net}\)) on the skydiver is zero (since \(F_{net} = ma\), and at terminal velocity, \(a = 0\)). Before terminal velocity, the skydiver accelerates downward, so \(F_{net}\) (gravity - air resistance) is downward and non - zero. As time increases towards 60 seconds, air resistance increases, reducing \(F_{net}\).

Step2: Analyze Each Position

• Position A (\(t = 0\)): At the initial state, air resistance is approximately zero (since velocity is zero). So \(F_{net}\) is equal to the force of gravity (\(F_g\)) and is the largest among the four positions (before terminal velocity).
• Position B (\(t = 15\) s): The skydiver has some velocity, so air resistance (\(F_{air}\)) is non - zero but less than \(F_g\). Thus, \(F_{net}=F_g - F_{air}\) is still downward and has a magnitude less than at \(t = 0\) but greater than at \(t = 45\) s.
• Position C (\(t = 45\) s): Air resistance has increased more. \(F_{net}=F_g - F_{air}\) has a smaller magnitude than at \(t = 15\) s, as the skydiver is closer to terminal velocity (but not yet there, so \(F_{net}

eq0\)).

• Position D (\(t = 60\) s): At terminal velocity, \(F_{net}=0\) (because \(F_g=F_{air}\) and they act in opposite directions, canceling each other out).

Answer:

• Position A: Largest non - zero net force (downward).
• Position B: Net force magnitude less than A, greater than C (downward).
• Position C: Net force magnitude less than B, greater than 0 (downward).
• Position D: Net force magnitude is 0.