Question
1. A site-seeing hot air balloon is 400 feet above an elephant. An observer on the ground sees the balloon from an angle of 5°. How far away from the elephant is the observer? © Success Academy Charter Schools 1
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3. A diver strikes the surface of the water in a pool at an angle of 80° from horizontal while traveling at a speed of 14 m/s. What is the horizontal velocity of the diver upon striking the water? © Success Academy Charter Schools 3
4. A shadow is 3 times longer than the object casting it. What angle is the Sun at?
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3. A diver strikes the surface of the water in a pool at an angle of 80° from horizontal while traveling at a speed of 14 m/s. What is the horizontal velocity of the diver upon striking the water? © Success Academy Charter Schools 3
4. A shadow is 3 times longer than the object casting it. What angle is the Sun at?
Answers
GPT-4o mini
### Problem 1: Hot Air Balloon and Observer
To find out how far away from the elephant the observer is, we can use the tangent function from trigonometry. The height of the hot air balloon above the elephant is 400 feet, and the angle of elevation from the observer to the balloon is 5°.
Let \( d \) be the distance from the observer to the elephant. We can set up the following equation using the tangent of the angle:
\[
\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{height}}{d}
\]
Given:
- angle = 5°
- height = 400 feet
We can rearrange the equation to solve for \( d \):
\[
d = \frac{\text{height}}{\tan(\text{angle})} = \frac{400}{\tan(5°)}
\]
Calculating \( \tan(5°) \):
\[
\tan(5°) \approx 0.0875 \quad (\text{using a calculator})
\]
Now substitute this value into the equation for \( d \):
\[
d = \frac{400}{0.0875} \approx 4571.43 \text{ feet}
\]
Thus, the observer is approximately **4571.43 feet** away from the elephant.
### Problem 3: Diver Striking Water
To find the horizontal velocity of the diver, we can use the cosine function, which relates the angle of the velocity vector to its horizontal component.
Given:
- speed of the diver = 14 m/s
- angle = 80°
The horizontal velocity (\( V_x \)) can be calculated with:
\[
V_x = V \cdot \cos(\theta)
\]
where \( V \) is the total speed and \( \theta \) is the angle from the horizontal.
Calculating \( \cos(80°) \):
\[
\cos(80°) \approx 0.1736 \quad (\text{using a calculator})
\]
Now substituting in:
\[
V_x = 14 \cdot \cos(80°) = 14 \cdot 0.1736 \approx 2.43 \text{ m/s}
\]
So, the horizontal velocity of the diver upon striking the water is approximately **2.43 m/s**.
### Problem 4: Shadow Length and Sun Angle
If a shadow is 3 times longer than the object casting it, we can denote:
- Let \( h \) be the height of the object.
- The length of the shadow = \( 3h \).
Using the tangent function, where \( \theta \) is the angle of elevation of the sun, we can set up the equation:
\[
\tan(\theta) = \frac{\text{height}}{\text{shadow length}} = \frac{h}{3h} = \frac{1}{3}
\]
To find the angle \( \theta \):
\[
\theta = \tan^{-1}\left( \frac{1}{3} \right)
\]
Calculating this:
\[
\theta \approx 18.43° \quad (\text{using a calculator})
\]
Thus, the angle of the sun is approximately **18.43 degrees**.
To find out how far away from the elephant the observer is, we can use the tangent function from trigonometry. The height of the hot air balloon above the elephant is 400 feet, and the angle of elevation from the observer to the balloon is 5°.
Let \( d \) be the distance from the observer to the elephant. We can set up the following equation using the tangent of the angle:
\[
\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\text{height}}{d}
\]
Given:
- angle = 5°
- height = 400 feet
We can rearrange the equation to solve for \( d \):
\[
d = \frac{\text{height}}{\tan(\text{angle})} = \frac{400}{\tan(5°)}
\]
Calculating \( \tan(5°) \):
\[
\tan(5°) \approx 0.0875 \quad (\text{using a calculator})
\]
Now substitute this value into the equation for \( d \):
\[
d = \frac{400}{0.0875} \approx 4571.43 \text{ feet}
\]
Thus, the observer is approximately **4571.43 feet** away from the elephant.
### Problem 3: Diver Striking Water
To find the horizontal velocity of the diver, we can use the cosine function, which relates the angle of the velocity vector to its horizontal component.
Given:
- speed of the diver = 14 m/s
- angle = 80°
The horizontal velocity (\( V_x \)) can be calculated with:
\[
V_x = V \cdot \cos(\theta)
\]
where \( V \) is the total speed and \( \theta \) is the angle from the horizontal.
Calculating \( \cos(80°) \):
\[
\cos(80°) \approx 0.1736 \quad (\text{using a calculator})
\]
Now substituting in:
\[
V_x = 14 \cdot \cos(80°) = 14 \cdot 0.1736 \approx 2.43 \text{ m/s}
\]
So, the horizontal velocity of the diver upon striking the water is approximately **2.43 m/s**.
### Problem 4: Shadow Length and Sun Angle
If a shadow is 3 times longer than the object casting it, we can denote:
- Let \( h \) be the height of the object.
- The length of the shadow = \( 3h \).
Using the tangent function, where \( \theta \) is the angle of elevation of the sun, we can set up the equation:
\[
\tan(\theta) = \frac{\text{height}}{\text{shadow length}} = \frac{h}{3h} = \frac{1}{3}
\]
To find the angle \( \theta \):
\[
\theta = \tan^{-1}\left( \frac{1}{3} \right)
\]
Calculating this:
\[
\theta \approx 18.43° \quad (\text{using a calculator})
\]
Thus, the angle of the sun is approximately **18.43 degrees**.