Explanation Given: The given expression is x = a t y = b t − 16 t 2 Calculation: As d y d x is the velocity of object at any point of time. Now, to find the derivative of the given equation so that to find the value of d y d x Thus, x = a t x ′ ( t ) = d x d t x ′ ( t ) = a y = b t − 16 t 2 y ′ ( t ) = d y d t y ′ ( t ) = b − 32 t Then, the value of d y d x is d y d x = d y d t d x d t d y d x = b − 32 t a At t = 0 d y d x is d y d x = b − 32 × 0 a d y d x = b a Since, d y d x

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ISBN: 9781319055905

Author: FRANZOSA

Publisher: VST

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Chapter 11.1, Problem 77E

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The projectile is launched at an angle θ=tan−1(ba) and lands at a distance of ab16 from the origin.

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Chapter 11.1, Problem 76E

Chapter 11.1, Problem 78E

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Chapter 11 Solutions

EBK CALCULUS: EARLY TRANSCENDENTALS

Ch. 11.1 - Prob. 1PQ Ch. 11.1 - Prob. 2PQ Ch. 11.1 - Prob. 3PQ Ch. 11.1 - Prob. 4PQ Ch. 11.1 - Prob. 5PQ Ch. 11.1 - Prob. 6PQ Ch. 11.1 - Prob. 7PQ Ch. 11.1 - Prob. 1E Ch. 11.1 - Prob. 2E Ch. 11.1 - Prob. 3E

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To proof: The projectile is launched at an angle θ = tan − 1 ( b a ) and lands at a distance of a b 16 from the origin.

Solution Summary: The author explains how the projectile is launched at an angle and lands at a distance of ab16 from the origin.

To proof:

The projectile is launched at an angle θ=tan−1(ba) and lands at a distance of ab16 from the origin.

Want to see the full answer?

Chapter 11 Solutions

To proof: The projectile is launched at an angle θ = tan − 1 ( b a ) and lands at a distance of a b 16 from the origin.

Solution Summary: The author explains how the projectile is launched at an angle and lands at a distance of ab16 from the origin.

To proof: The projectile is launched at an angle θ=tan−1(ba) and lands at a distance of ab16 from the origin.

Want to see the full answer?

Chapter 11 Solutions

To proof:

The projectile is launched at an angle θ=tan−1(ba) and lands at a distance of ab16 from the origin.