(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: °
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter9: Quadratic Functions And Equations
Section9.6: Solving Quadratic Equations By Using The Quadratic Formula
Problem 64PFA
Related questions
Question
See attached.
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd76adacd-674b-4971-a3ab-f7962e1f5ed3%2F26bd233a-c801-4037-b781-62e37b1ae7a4%2F4kb89wh_processed.png&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd76adacd-674b-4971-a3ab-f7962e1f5ed3%2F26bd233a-c801-4037-b781-62e37b1ae7a4%2Fnn8a0xn_processed.png&w=3840&q=75)
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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