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(a) (b) (c) How many seconds does it take for the target to reach its maximum height? Round your answer to the nearest hundredth. Time to reach maximum height: seconds Using your answer from part (a), find the maximum height (in meters) the target reaches. Round your answer to the nearest hundredth. Maximum height: m Using your answer from part (b), find the angle of depression (in degrees) of the archer's line of sight when the target reaches its maximum height. Round your answer to the nearest hundredth. Angle of depression: °

(a) (b) (c) How many seconds does it take for the target to reach its maximum height? Round your answer to the nearest hundredth. Time to reach maximum height: seconds Using your answer from part (a), find the maximum height (in meters) the target reaches. Round your answer to the nearest hundredth. Maximum height: m Using your answer from part (b), find the angle of depression (in degrees) of the archer's line of sight when the target reaches its maximum height. Round your answer to the nearest hundredth. Angle of depression: °

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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A disc-shaped target is launched straight up into the sky. The target's height h (in meters) above the launcher after t seconds is given by the following function. h(t) = − 4.9t² + 14.6t An archer is high up on a nearby tower, 15 meters above the target launcher. She tracks the target using a sight as it travels into the sky. The horizontal distance between the archer's eyes and the target's vertical path is 18 meters, as shown below. (The figure is not drawn to scale.) 18 m 15 m
Transcribed Image Text:A disc-shaped target is launched straight up into the sky. The target's height h (in meters) above the launcher after t seconds is given by the following function. h(t) = − 4.9t² + 14.6t An archer is high up on a nearby tower, 15 meters above the target launcher. She tracks the target using a sight as it travels into the sky. The horizontal distance between the archer's eyes and the target's vertical path is 18 meters, as shown below. (The figure is not drawn to scale.) 18 m 15 m
Answer the following. (a) How many seconds does it take for the target to reach its maximum height? Round your answer to the nearest hundredth. Time to reach maximum height: seconds (b) (c) Using your answer from part (a), find the maximum height (in meters) the target reaches. Round your answer to the nearest hundredth. Maximum height: m Using your answer from part (b), find the angle of depression (in degrees) of the archer's line of sight when the target reaches its maximum height. Round your answer to the nearest hundredth. Angle of depression: O
Transcribed Image Text:Answer the following. (a) How many seconds does it take for the target to reach its maximum height? Round your answer to the nearest hundredth. Time to reach maximum height: seconds (b) (c) Using your answer from part (a), find the maximum height (in meters) the target reaches. Round your answer to the nearest hundredth. Maximum height: m Using your answer from part (b), find the angle of depression (in degrees) of the archer's line of sight when the target reaches its maximum height. Round your answer to the nearest hundredth. Angle of depression: O
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