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An arrow is shot upward, with an initial velocity of 90 meters per second, at an angle of 34° with respect to the horizontal. The arrow is shot from a height of 6 meters above the ground. The horizontal distance x from the starting point and the height y above the ground of the arrow t seconds after it is shot are given by the parametric equations below. = (v₁ cose)t y=−4.9t²+(v sin0)t+h X Here vo is the initial velocity, is the initial angle with respect to the horizontal, and h is the initial height. Use the equations to answer the following questions.

An arrow is shot upward, with an initial velocity of 90 meters per second, at an angle of 34° with respect to the horizontal. The arrow is shot from a height of 6 meters above the ground. The horizontal distance x from the starting point and the height y above the ground of the arrow t seconds after it is shot are given by the parametric equations below. = (v₁ cose)t y=−4.9t²+(v sin0)t+h X Here vo is the initial velocity, is the initial angle with respect to the horizontal, and h is the initial height. Use the equations to answer the following questions.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter11: Topics From Analytic Geometry
Section: Chapter Questions
Problem 10DE
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An arrow is shot upward, with an initial velocity of 90 meters per second, at an angle of 34° with respect to the horizontal. The arrow is shot from a height of 6 meters above the ground. The horizontal distance x from the starting point and the height y above the ground of the arrow t seconds after it is shot are given by the parametric equations below. = (v₁ cos 0)t y = −4.9t²+(v₁ sin0)t+h X Here Vo is the initial velocity, is the initial angle with respect to the horizontal, and h is the initial height. Use the equations to answer the following questions.
Transcribed Image Text:An arrow is shot upward, with an initial velocity of 90 meters per second, at an angle of 34° with respect to the horizontal. The arrow is shot from a height of 6 meters above the ground. The horizontal distance x from the starting point and the height y above the ground of the arrow t seconds after it is shot are given by the parametric equations below. = (v₁ cos 0)t y = −4.9t²+(v₁ sin0)t+h X Here Vo is the initial velocity, is the initial angle with respect to the horizontal, and h is the initial height. Use the equations to answer the following questions.
(a) When does the arrow reach its maximum height? Do not round any intermediate computations. Round your answer to the nearest hundredth. seconds (b) What is the maximum height of the arrow? Round your answer to the nearest tenth. meters X Ś FR
Transcribed Image Text:(a) When does the arrow reach its maximum height? Do not round any intermediate computations. Round your answer to the nearest hundredth. seconds (b) What is the maximum height of the arrow? Round your answer to the nearest tenth. meters X Ś FR
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