Program Explanation Given information: A crossbow bolt shot straight upward from the ground. y 0 = 108.47 m The initial velocity of the bolt is 0 . v 0 = 0 m / s The velocity function for downward motion is v ( t ) = g ρ tanh ( C 2 − t ρ g ) where C 2 = tanh − 1 ( v 0 ρ g ) and y ( 0 ) = y 0 . The upward position function y ( t ) = y 0 − 1 ρ ln | cosh ( C 2 − t ρ g ) cosh C 2 | The value of ρ is 0.0011 and value of g is 9.8 m / s 2 . Explanation: Substitute the value of ρ , g , v 0 in C 2 and solve as follows: C 2 = tan − 1 ( 0 0.0011 9.8 ) = tan − 1 ( 0 ) = 0 Use upward position function y ( t ) and solve for y ( t ) = 0 as follows: 108.47 − 1 0.0011 ln | cosh ( 0 − t ( 0.0011 ) ( 9.8 ) ) cosh 0 | = 0 1 0

5th Edition

ISBN: 9780321974235

Author: Calvis

Publisher: PEARSON CUSTOM PUB.(CONSIGNMENT)

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Longest Common Subsequence

The Longest Common Subsequence (LCS) is defined as the longest subsequence that is shared by all of the given sequences, assuming that the subsequence's elements do not have to be in consecutive locations in the original sequences. It is used in the area…

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Chapter 2.3, Problem 18P

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Program Description: Purpose of the problem is to show that bolt hits the ground about 4.80 s later with an impact speed of about 43.49 m/s , if the bold is dropped v0=0 from a height of y0=108.47m .

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Chapter 2.3, Problem 17P

Chapter 2.3, Problem 19P

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Program Description: Purpose of the problem is to show that bolt hits the ground about 4.80 s later with an impact speed of about 43.49 m / s , if the bold is dropped v 0 = 0 from a height of y 0 = 108.47 m .

Solution Summary: The author explains that a crossbow bolt hits the ground about 4.80 s later with an impact speed of about 43.49

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Program Description: Purpose of the problem is to show that bolt hits the ground about 4.80 s later with an impact speed of about 43.49 m/s , if the bold is dropped v0=0 from a height of y0=108.47m .

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Chapter 2 Solutions