A helicopter hovers 40 ft above the ground. Then the helicopter climbs at a rate of 21 ft/s. How to write a rule that represents the helicopter's height h above the ground as a function of time t. What is the helicopter's height after 45 s?

y1i2de1o64w

y1i2de1o64w

Answered question

2022-12-31

A helicopter hovers 40 ft above the ground. Then the helicopter climbs at a rate of 21 ft/s. How to write a rule that represents the helicopter's height h above the ground as a function of time t. What is the helicopter's height after 45 s?

Answer & Explanation

hemelbol26w

hemelbol26w

Beginner2023-01-01Added 9 answers

The height of the helicopter at launch is 40. Since this won't change, the height equation will use a constant to represent it.
This can be shown as an additional 21 feet because the helicopter climbs at a speed of 21 feet per second. When t=1, an additional 21 feet is being added to the helicopter's height. When t=2, an additional 42 feet is being added.
Then:
h=40+21t
After 45 seconds:
h=40+21(45)
h=40+945
h=985feet

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