a hot air balloon is ascending at a constant angle of elevation from point on the ground. the horizontal distance from the point of observation to the balloon is represented by x(t)=5t, where t is time in minutes.
task 1- if the angle of elevation is 45 degree and the balloons height above the ground is given by h(t)=5t tan45, derive the equation of the path of the balloon
task 2- suppose the angle of elevation increases to 60 degrees. recalculate the balloon's height equation and graph both paths on the same set of axes
task 3- determine the time when the balloon's height reaches 100m for each angles. discuss how the angle of elevation affects the rate of ascent.

Let's break down and solve each task one by one.

### **Task 1: Deriving the Equation of the Path of the Balloon (Angle of Elevation = 45°)**

Given:
- The horizontal distance from the point of observation to the balloon is \( x(t) = 5t \), where \( t \) is time in minutes.
- The angle of elevation is 45°.
- The height of the balloon above the ground is given by \( h(t) = 5t \tan 45^\circ \).

**Step 1: Simplify the height function.**

Since \( \tan 45^\circ = 1 \):
\[
h(t) = 5t \times 1 = 5t
\]

**Step 2: Derive the equation of the path.**

The path of the balloon is the relationship between the height \( h(t) \) and the horizontal distance \( x(t) \).

Since \( h(t) = 5t \) and \( x(t) = 5t \), we can express the height in terms of the horizontal distance:

\[
h = x
\]

This equation \( h = x \) represents the path of the balloon when the angle of elevation is 45°.

### **Task 2: Recalculate the Balloon's Height Equation (Angle of Elevation = 60°) and Graph Both Paths**

Given:
- The angle of elevation increases to 60°.

**Step 1: Derive the new height function.**

For the new angle of elevation:
\[
h(t) = 5t \tan 60^\circ
\]

Since \( \tan 60^\circ = \sqrt{3} \):
\[
h(t) = 5t \sqrt{3}
\]

**Step 2: Graph both paths.**

For \( \theta = 45^\circ \):
\[
h = x
\]

For \( \theta = 60^\circ \):
\[
h = x \sqrt{3}
\]

The graph of these two equations on the same set of axes would show two straight lines, with the line for \( \theta = 60^\circ \) (i.e., \( h = x \sqrt{3} \)) being steeper than the line for \( \theta = 45^\circ \) (i.e., \( h = x \)).

### **Task 3: Determine the Time When the Balloon's Height Reaches 100 m for Each Angle and Discuss**

**Step 1: Solve for \( t \) when \( h(t) = 100 \) m for each angle.**

- **For \( \theta = 45^\circ \):**
  \[
  5t = 100
  \]
  \[
  t = \frac{100}{5} = 20 \text{ minutes}
  \]

- **For \( \theta = 60^\circ \):**
  \[
  5t \sqrt{3} = 100
  \]
  \[
  t = \frac{100}{5\sqrt{3}} = \frac{20}{\sqrt{3}} \approx 11.55 \text{ minutes}
  \]

**Step 2: Discuss the effect of the angle of elevation on the rate of ascent.**

The time it takes for the balloon to reach a height of 100 meters is shorter when the angle of elevation is greater. This is because a steeper angle (larger angle of elevation) means the balloon ascends more vertically, covering the same vertical distance in less time.

### **Summary:**
- For \( \theta = 45^\circ \), the height equation is \( h = x \).
- For \( \theta = 60^\circ \), the height equation is \( h = x \sqrt{3} \).
- The balloon reaches 100 m in 20 minutes at \( \theta = 45^\circ \) and in about 11.55 minutes at \( \theta = 60^\circ \).
- A larger angle of elevation leads to a faster ascent.

---

Do you want further details or have any questions?

### **Relative Questions:**
1. How would the path change if the angle of elevation were 30°?
2. What would be the effect on the balloon's path if the rate of horizontal movement changed?
3. Can we derive a general formula for the height as a function of angle and time?
4. How would the situation change if the balloon's ascent was not at a constant rate?
5. What would be the balloon's path if the angle of elevation varied with time?

### **Tip:**
When solving trigonometric problems, it’s helpful to remember key angle values and their corresponding trigonometric functions.

Explore the path of a hot air balloon ascending at varying angles of elevation. Solve for the balloon's height equations at 45° and 60°, graph their paths, and determine ascent times to 100m. Learn how elevation angles affect ascent rates.

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