Explanation Given: The motion of an object moving vertically through the air is governed by the equation, d 2 y d t 2 = − g − g V 2 d y d t | d y d t | Here, y is the upward vertical displacement and V is a constant called the terminal speed. Calculation: Substitute 32 ft / sec 2 for g , 50 ft / sec for V . d 2 y d t 2 = − 32 − 32 2500 d y d t | d y d t | ... ( 1 ) Assume v = d y d t so, y ′ = v y ″ = v ′ Substitute v ′ for d y d t in equation ( 1 ) . v ′ = − 32 − 16 1250 v | v | Substitute 0 for v ′ in above equation

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Chapter 5.4, Problem 34E

To determine

To sketch:

The trajectories in the -phase plane for −100≤y≤100, −100≤v≤100 starting from y=0, v=−75,−50,−25,0,25,50 and, 75 ft/sec and find terminal velocity.

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Chapter 5.4, Problem 32E

Chapter 5.4, Problem 35E

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The trajectories in the -phase plane for − 100 ≤ y ≤ 100 , − 100 ≤ v ≤ 100 starting from y = 0 , v = − 75 , − 50 , − 25 , 0 , 25 , 50 and, 75 ft / sec and find terminal velocity.

Solution Summary: The author illustrates the trajectories in the phase plane for -100 le yle 100 and the terminal velocity of 50.

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To sketch:

The trajectories in the -phase plane for −100≤y≤100, −100≤v≤100 starting from y=0, v=−75,−50,−25,0,25,50 and, 75 ft/sec and find terminal velocity.

Want to see the full answer?

Chapter 5 Solutions

The trajectories in the -phase plane for − 100 ≤ y ≤ 100 , − 100 ≤ v ≤ 100 starting from y = 0 , v = − 75 , − 50 , − 25 , 0 , 25 , 50 and, 75 ft / sec and find terminal velocity.

Solution Summary: The author illustrates the trajectories in the phase plane for -100 le yle 100 and the terminal velocity of 50.

Concept explainers

Videos

To sketch: The trajectories in the -phase plane for −100≤y≤100, −100≤v≤100 starting from y=0, v=−75,−50,−25,0,25,50 and, 75 ft/sec and find terminal velocity.

Want to see the full answer?

Chapter 5 Solutions

To sketch:

The trajectories in the -phase plane for −100≤y≤100, −100≤v≤100 starting from y=0, v=−75,−50,−25,0,25,50 and, 75 ft/sec and find terminal velocity.