A balloon leaves from a point P rises at a uniform speed. After 6 minutes, an observer situated at a distance of `450 sqrt""3` metres from point P observes that angle of elevation of the balloon is `60^(@)` Assume that point of observation and point P are on the same level. What is the speed (in m/s) of the balloon?

Options

A. `4 . 25`
B. `3 . 75`
C. `4 . 5`
D. `3 . 45`

Answer

To solve the problem step by step, we will follow the given information and apply trigonometric principles.

### Step 1: Understand the Problem
We have a balloon rising from point P, and after 6 minutes, an observer at a distance of \( 450\sqrt{3} \) meters from point P observes the balloon at an angle of elevation of \( 60^\circ \). We need to find the speed of the balloon in meters per second.

### Step 2: Set Up the Right Triangle
When the observer looks at the balloon, we can form a right triangle where:
- The horizontal distance from point P to the observer is \( 450\sqrt{3} \) meters.
- The height of the balloon at that moment is \( H \) meters.
- The angle of elevation is \( 60^\circ \).

### Step 3: Use the Tangent Function
From the right triangle, we can use the tangent of the angle of elevation:
\[ \tan(60^\circ) = \frac{H}{450\sqrt{3}} \]
We know that \( \tan(60^\circ) = \sqrt{3} \). Thus, we can write:
\[ \sqrt{3} = \frac{H}{450\sqrt{3}} \]

### Step 4: Solve for Height \( H \)
Rearranging the equation gives:
\[ H = 450\sqrt{3} \cdot \sqrt{3} = 450 \cdot 3 = 1350 \text{ meters} \]

### Step 5: Calculate the Speed of the Balloon
The speed of the balloon is defined as the height gained per unit time. Since the balloon rises to a height of \( 1350 \) meters in \( 6 \) minutes, we first convert the time into seconds:
\[ 6 \text{ minutes} = 6 \times 60 = 360 \text{ seconds} \]
Now, we can calculate the speed:
\[ \text{Speed} = \frac{H}{\text{Time}} = \frac{1350 \text{ meters}}{360 \text{ seconds}} = 3.75 \text{ m/s} \]

### Conclusion
The speed of the balloon is \( 3.75 \) m/s.

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