Explanation Given: The astronaut hits his golf shot on the moon with an initial speed of 35 m/s at angle of 30 ° . Calculation: Since the acceleration due to gravity is 1.6 m/s , the velocity vector can be written as v ( t ) = ∫ a ( t ) d t = 〈 0 , − 1.6 t 〉 + C 0 ........ (1) The ball was hit with speed of 35 m/s at an angle of 30 ° . So, it gives v ( 0 ) = 〈 35 cos θ , 35 sin θ 〉 . From (1), v ( 0 ) = 〈 0 , − 1.6 ⋅ 0 〉 + C 0 v ( 0 ) = 〈 0 , 0 〉 + C 0 v ( 0 ) = C 0 C 0 = 〈 35 cos θ , 35 sin θ 〉 Therefore, the equation (1) becomes, v ( t ) = 〈 0 , − 1.6 t 〉 + 〈 35 cos θ , 35 sin θ 〉 = 〈 35 cos θ , 35 sin θ − 1.6 t 〉 Then, r ( t ) = ∫ 〈 35 cos θ , 35 sin θ − 1.6 t 〉 = 〈 35 cos θ t + K 1 , 35 sin θ t − 0.8 t 2 + K 2 〉 At t = 0 , r ( t ) = 〈 0 , 0 〉

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

4th Edition

ISBN: 9781319312886

Author: Rogawski, Jon, Adams, Colin, FRANZOSA, Robert

Publisher: W. H. Freeman

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Chapter 13.5, Problem 21E

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The distance travelled by the ball when the astronaut hits his golf shot on the moon with an initial speed of 35 m/s at angle of 30°.

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Chapter 13.5, Problem 20E

Chapter 13.5, Problem 22E

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Chapter 13 Solutions

Loose-Leaf Version for Calculus: Early Transcendentals Single Variable

Ch. 13.1 - Prob. 1PQ Ch. 13.1 - Prob. 2PQ Ch. 13.1 - Prob. 3PQ Ch. 13.1 - Prob. 4PQ Ch. 13.1 - Prob. 5PQ Ch. 13.1 - Prob. 6PQ Ch. 13.1 - Prob. 1E Ch. 13.1 - Prob. 2E Ch. 13.1 - Prob. 3E Ch. 13.1 - Prob. 4E

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The distance travelled by the ball when the astronaut hits his golf shot on the moon with an initial speed of 35 m/s at angle of 30 ° .

Solution Summary: The astronaut hits his golf shot on the moon with an initial speed of 35 m/s at an angle of 30°. The velocity vector can be written as v(t)=langle

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The distance travelled by the ball when the astronaut hits his golf shot on the moon with an initial speed of 35 m/s at angle of 30°.

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Chapter 13 Solutions