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: Let us consider the given differential equation. 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 Thus, the solution is given by v ( t ) = 1 e k m t ( − 9 To determine (b) To prove: The projectile’s velocity is zero at time t * = m k ln ( v 0 k 9.8 m + 1 ) To determine (c) The height y ( t ) . To determine (d) To prove: y ( t * ) = m k v 0 − 9.8 m 2 ln ( v 0 k 9.8 m + 1 ) k 2

ACHIEVE FOR CALCULUS 4 TERM >CSI CUSTOM<

21st Edition

ISBN: 9781319438333

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 9.2, Problem 26E

To determine

(a)

The velocity v(t).

To determine

(b)

To prove:

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

To determine

(c)

The height y(t).

To determine

(d)

To prove:

y(t*)=mkv0−9.8m2ln(v0k9.8m+1)k2

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Chapter 9.2, Problem 25E

Chapter 9.3, Problem 1PQ

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Chapter 9 Solutions

ACHIEVE FOR CALCULUS 4 TERM >CSI CUSTOM<

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