a. Explanation Given information: A projectile of mass m kg travels straight up from ground level with initial velocity v 0 m/s. Velocity v satisfies v ' = − 9.8 − k m v ( t ) . Theorem used: The general solution of y ' + P ( x ) y = Q ( x ) is y = 1 α ( x ) ( ∫ α ( x ) Q ( x ) d x ) where α ( x ) , an integrating factor: α ( x ) = e ∫ P ( x ) d x Calculation: v ' = − 9.8 − k m v ( t ) ⇒ v ' + k m v ( t ) = − 9.8 It is of the form y ' + P ( x ) y = Q ( x ) with P ( x ) = k m , Q ( x ) = − 9.8 , then the solution is v ( t ) = 1 α ( t ) ( ∫ α ( t ) ( − 9.8 ) d t ) where α ( t ) = e ∫ P ( t ) d t = e ∫ k m d t = e k m t v ( t ) = 1 e k m t ( − 9.8 ∫ e k m t d t ) v ( t ) = 1 e k m t ( − 9 b. To determine To prove: The projectile’s velocity is zero at time t * = m k ln ( v 0 k 9.8 m + 1 ) c. To determine The height y ( t ) d. To determine To prove : y ( t * ) = m k v 0 − 9.8 m 2 ln ( v 0 k 9.8 m + 1 ) k 2

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Chapter 10.2, Problem 26E

To determine

To prove: The velocity v(t)

To determine

To prove:The projectile’s velocity is zero at time t*=mkln(v0k9.8m+1)

To determine

The height y(t)

To determine

To prove: y(t*)=mkv0−9.8m2ln( v 0 k 9.8m+1)k2

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Chapter 10.2, Problem 25E

Chapter 10.3, Problem 1PQ

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To prove: The velocity v ( t )

Solution Summary: The author explains the general solution of y'=-9.8-kmv(t).

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To prove: The velocity v(t)

To prove:The projectile’s velocity is zero at time t*=mkln(v0k9.8m+1)

The height y(t)

To prove: y(t*)=mkv0−9.8m2ln( v 0 k 9.8m+1)k2

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Chapter 10 Solutions