Explanation Given information: Assume that for a projectile launched below is given, With a modest initial speed = v 0 Acceleration under gravity on earth= g Angle= θ Formula Used: Equations of motion s = u t + 1 2 a t 2 v 2 = u 2 + 2 a s v = u + a t Calculation: At the highest point of the path of the golf shot speed v = 0 . Consider the equation v = u + a t in the y → direction. Then, v = u + a t 0 = v 0 s i n θ − g t t = v 0 s i n θ g Consider the equation s = u t + 1 2 a t 2 in the x → direction. Then, s = u t + 1 2 a t 2 s = v 0 c o s θ . v 0 s i n θ g + 0 s = v 0 2 g s i n θ c o s θ Therefore, thetotal distance travelled is d = 2 s . d = 2 s d = 2 v 0 2 g s i n θ c o s θ d = v 0 2 g s i n 2 θ Now we need to differentiate d with respect to θ and find d ( d ) d θ

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Chapter 14.5, Problem 23E

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Show that the distance travelled is v02gsin2θ .And the maximum distance d is attained at θ=π4 .

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Chapter 14.5, Problem 22E

Chapter 14.5, Problem 24E

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Chapter 14 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

Ch. 14.1 - Prob. 1PQ Ch. 14.1 - Prob. 2PQ Ch. 14.1 - Prob. 3PQ Ch. 14.1 - Prob. 4PQ Ch. 14.1 - Prob. 5PQ Ch. 14.1 - Prob. 6PQ Ch. 14.1 - Prob. 1E Ch. 14.1 - Prob. 2E Ch. 14.1 - Prob. 3E Ch. 14.1 - Prob. 4E

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Show that the distance travelled is v 0 2 g s i n 2 θ .And the maximum distance d is attained at θ = π 4 .

Solution Summary: The author explains how the distance travelled is d=v_02gsin2's theta, and the maximum distance is at the highest point of the path of

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Show that the distance travelled is v02gsin2θ .And the maximum distance d is attained at θ=π4 .

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Chapter 14 Solutions