Explanation Given: The distance of shot is given as, d ( θ ) = 5.1 sin ( 2 θ ) [ 1 + 1 + 0.41 sin 2 θ ] ... ( 1 ) Approach: Product rule of differentiation . d d x ( x 1 x 2 ) = x 1 d d x ( x 2 ) + x 2 d d x ( x 1 ) Calculation: The release angle and maximum distance is calculated as follows. Apply the product rule of differentiation in the equation ( 1 ) . d d θ ( d ( θ ) ) = 5.1 sin ( 2 θ ) d d θ [ 1 + 1 + 0.41 sin 2 θ ] + [ 1 + 1 + 0.41 sin 2 θ ] d d θ ( 5.1 sin ( 2 θ ) ) d ( d ( θ ) ) d θ = 5.1 sin ( 2 θ ) d d θ [ 1 + 1 + 0.41 sin 2 θ ] + [ 1 + 1 + 0.41 sin 2 θ ] d d θ ( 5.1 sin ( 2 θ ) ) d ( d ( θ ) ) d θ = 5

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ISBN: 9780321964038

Author: GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.

Publisher: Pearson Addison Wesley,

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Chapter 6.2, Problem 19E

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The value of the release angle that maximizes the distance traveled by the shot and corresponding maximum distance.

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Chapter 6.2, Problem 17E

Chapter 6.2, Problem 20E

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The value of the release angle that maximizes the distance traveled by the shot and corresponding maximum distance.

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To find: The value of the release angle that maximizes the distance traveled by the shot and corresponding maximum distance.

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Chapter 6 Solutions

To find:

The value of the release angle that maximizes the distance traveled by the shot and corresponding maximum distance.