ariants/512235/take/13/ SKETCHPAD Question 13 A plane takes off from an airport at an angle of 14.5°. If the plane is traveling at a velocity of 220 feet per second, how high off of the ground is the plane at 10 seconds? THIN V BLACK N B ☐ M
ariants/512235/take/13/ SKETCHPAD Question 13 A plane takes off from an airport at an angle of 14.5°. If the plane is traveling at a velocity of 220 feet per second, how high off of the ground is the plane at 10 seconds? THIN V BLACK N B ☐ M
Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE:
1. Give the measures of the complement and the supplement of an angle measuring 35°.
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![### Question 13
A plane takes off from an airport at an angle of 14.5°. If the plane is traveling at a velocity of 220 feet per second, how high off of the ground is the plane at 10 seconds?
**Explanation:**
- **Given:**
- Angle of ascent: 14.5°
- Velocity of the plane: 220 feet per second
- Time elapsed: 10 seconds
To solve this problem, you can use basic trigonometry. The height \( h \) the plane reaches can be found using the sine function in a right-angled triangle where the angle is 14.5° and the hypotenuse is the distance traveled by the plane.
1. **Calculate the Hypotenuse:**
\[
\text{Distance} = \text{Velocity} \times \text{Time}
\]
\[
\text{Distance} = 220 \, \text{feet/second} \times 10 \, \text{seconds} = 2200 \, \text{feet}
\]
2. **Find the Opposite Side (Height) using Sine Function:**
\[
\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}
\]
\[
\sin(14.5°) = \frac{h}{2200 \, \text{feet}}
\]
3. **Solve for \( h \):**
\[
h = 2200 \, \text{feet} \times \sin(14.5°)
\]
Using a calculator to find \(\sin(14.5°)\):
\[
\sin(14.5°) \approx 0.25038
\]
Thus,
\[
h = 2200 \times 0.25038 \approx 550.836 \, \text{feet}
\]
So, the plane is approximately 550.836 feet above the ground after 10 seconds.
**Note:** To further visualize, you may include a right-angled triangle diagram with the given angle, velocity vector (as hypotenuse), and computed height.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F075db94c-9df1-4bd0-8bd8-7247d457f29a%2F3dea742f-ed0b-40ab-adf1-6b4f11827da0%2Ftvnwtd_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 13
A plane takes off from an airport at an angle of 14.5°. If the plane is traveling at a velocity of 220 feet per second, how high off of the ground is the plane at 10 seconds?
**Explanation:**
- **Given:**
- Angle of ascent: 14.5°
- Velocity of the plane: 220 feet per second
- Time elapsed: 10 seconds
To solve this problem, you can use basic trigonometry. The height \( h \) the plane reaches can be found using the sine function in a right-angled triangle where the angle is 14.5° and the hypotenuse is the distance traveled by the plane.
1. **Calculate the Hypotenuse:**
\[
\text{Distance} = \text{Velocity} \times \text{Time}
\]
\[
\text{Distance} = 220 \, \text{feet/second} \times 10 \, \text{seconds} = 2200 \, \text{feet}
\]
2. **Find the Opposite Side (Height) using Sine Function:**
\[
\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}
\]
\[
\sin(14.5°) = \frac{h}{2200 \, \text{feet}}
\]
3. **Solve for \( h \):**
\[
h = 2200 \, \text{feet} \times \sin(14.5°)
\]
Using a calculator to find \(\sin(14.5°)\):
\[
\sin(14.5°) \approx 0.25038
\]
Thus,
\[
h = 2200 \times 0.25038 \approx 550.836 \, \text{feet}
\]
So, the plane is approximately 550.836 feet above the ground after 10 seconds.
**Note:** To further visualize, you may include a right-angled triangle diagram with the given angle, velocity vector (as hypotenuse), and computed height.
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