Explanation The newton’s law of motion is given by d ( m v ) d t = F + ( v + u ) d m d t . The initial mass of the rocket is m 0 and rate at which mass is directing backward is d m d t = − b . The constant speed relative to the rocket is u = − c . The external force acting is F = − m g . Simplify the Newton’s law of motion as follows. d ( m v ) d t = F + ( v + u ) d m d t F = d ( m v ) d t − ( v + u ) d m d t F = m d v d t + v d m d t − v d m d t − u d m d t F = m d v d t − u d m d t Consider d m d t = − b and evaluate the value of m as follows. d m d t = − b ∫ d m = − ∫ b d t m = − | b | t + C Substitute the initial condition t = 0 and evaluate the value of C as follows. m ( 0 ) = − | b | ( 0 ) + C m 0 = C Therefore, the value of m is m = − | b | t + m 0 . Substitute m = − | b | t + m 0 in F = m d v d t − u d m d t as follows. F = ( − | b | t + m 0 ) d v d t − u ( − | b | ) − m g = ( m 0 − | b | t ) d v d t + u | b | − ( − | b | t + m 0 ) g = ( m 0 − | b | t ) d v d t + u | b | d v d t = − g + u | b | m 0 − | b | t Integrate and evaluate the value of v as follows

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Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differentia…

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Chapter 16, Problem 2AAE

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To show: The height of the rocket above the ground at the end of t seconds is y=c[t+m0−btblnm0−btm0]−12gt2.

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Chapter 16 Solutions

MYLAB MATH W/PEARSON ETEXT 18 WEEK CODE

Ch. 16.1 - In Exercises 1–4, match the differential equations...Ch. 16.1 - In Exercises 1–4, match the differential equations...Ch. 16.1 - In Exercises 1–4, match the differential equations...Ch. 16.1 - In Exercises 1–4, match the differential equations...Ch. 16.1 - In Exercises 5 and 6, copy the slope fields, and...Ch. 16.1 - In Exercises 5 and 6, copy the slope fields, and...Ch. 16.1 - Prob. 7E Ch. 16.1 - Prob. 8E Ch. 16.1 - Prob. 9E Ch. 16.1 - Prob. 10E

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The height of the rocket above the ground at the end of t seconds is y = c [ t + m 0 − b t b ln m 0 − b t m 0 ] − 1 2 g t 2 .

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To show: The height of the rocket above the ground at the end of t seconds is y=c[t+m0−btblnm0−btm0]−12gt2.

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