Explanation Given: Theprojectile is fired at an angle 45°with the horizontal at an initial speed of 150 feet/second. Formula Used: a T = v . a ‖ v ‖ i f r ( t ) = a i+b j+c k ||r(t)||= a 2 + b 2 + c 2 a n = ‖ v × a ‖ ‖ v ‖ Calculation: On referring to 52516-12.4-57E-A, we know that the tangential and normal components of acceleration are: a T = − 32 ( v 0 sin θ − 32 t ) v 0 2 cos 2 θ + ( v 0 sin θ − 32 t ) 2 and a N = 32 v 0 c o s θ v 0 2 cos 2 θ + ( v 0 sin θ − 32 t ) 2 have , v 0 = 150 f e e t / sec o n d θ = 45 ∘ a T = − 32 ( 150 2 − 32 t ) ( 150 ) 2 − 2 150 2 × 32 t + ( 32 t ) 2 = − 150 2 × 32 − ( 32 × 32 ) t 4 [ ( 75 ) 2 − 2 × 75 × 16 t + ( 16 ) 2 t 2 ] = − 150 × 16 2 − ( 32 × 32 ) t 2 [ ( 75 ) 2 − 2 × 75 × 16 t + ( 16 ) 2 t 2 ] = − 1697

Calculus: Early Transcendental Functions

7th Edition

ISBN: 9781337552516

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 12.4, Problem 58E

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To Calculate: The tangential and normal components of acceleration for a projectile fired at an angle 45° with the horizontal at an initial speed of 150 feet/second. Also find the components when the projectile is at its maximum height.

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Chapter 12.4, Problem 57E

Chapter 12.4, Problem 59E

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Calculus: Early Transcendental Functions

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The tangential and normal components of acceleration for a projectile fired at an angle 45° with the horizontal at an initial speed of 150 feet/second. Also find the components when the projectile is at its maximum height.

Solution Summary: The author calculates the tangential and normal components of acceleration for a projectile fired at an angle of 45° with the horizontal at the initial speed of 150 feet/second.

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To Calculate: The tangential and normal components of acceleration for a projectile fired at an angle 45° with the horizontal at an initial speed of 150 feet/second. Also find the components when the projectile is at its maximum height.

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Chapter 12 Solutions