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.
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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