Program Explanation Given information: An arrow is shot straight upward from the ground with an initial velocity of 160 ft / s . Arrow experiences gravity and deceleration v 2 800 due to air resistance. Explanation: The differential equation for the given system is written below. d v d t = − 32 − v 2 800 Here, constant gravitational acceleration is 32 ft / s 2 and deceleration due to air resistance is v 2 800 . Compare the above differential equation is in the form of upward motion d v d t = − g − ρ v 2 that has velocity function and position function as follows: v ( t ) = g ρ tan ( C 1 − t g ρ ) y ( t ) = y 0 + 1 ρ ln | cos ( C 1 − t g ρ ) cos C 1 | C 1 = tan − 1 ( v 0 ρ g ) Obtain the value of C 1 as follows: C 1 = tan − 1 ( 160 1 800 ⋅ 1 32 ) = tan − 1 ( 160 ⋅ 1 160 ) = tan − 1 ( 1 ) = π 4 Impose the condition that maximum height is attained when velocity function is zero

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Author: Calvis

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A naïve solution for the above-stated problem is to evaluate all of the possible combinations of different pieces of varying size to find the maximum revenue-generating combination. Although this method seems the easiest and quick to practice, it has exp…

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Chapter 2.3, Problem 20P

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Program Description: Purpose of the problem is to obtain the maximum height attained by the arrow, if arrow is shot straight upward from the ground with an initial velocity of 160ft/s .

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Chapter 2.3, Problem 19P

Chapter 2.3, Problem 21P

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Program Description: Purpose of the problem is to obtain the maximum height attained by the arrow, if arrow is shot straight upward from the ground with an initial velocity of 160 ft / s .

Solution Summary: The author explains that the problem is to obtain the maximum height attained by the arrow if it is shot straight upward from the ground.

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Program Description: Purpose of the problem is to obtain the maximum height attained by the arrow, if arrow is shot straight upward from the ground with an initial velocity of 160ft/s .

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Chapter 2 Solutions