An airplane takes off at an angle `30^(@)` with the horizontal ground treveling at the speed of `180km//h`. If it continuous to fly with the same velocity in the same direction, how long will it take to reach an altitude of `9km` above the ground?

Options

A. `5` minutes
B. `6` minutes
C. `8` minutes
D. `9` minutes

Explanation

To solve the problem, we need to determine how long it will take for the airplane to reach an altitude of 9 km while flying at an angle of 30 degrees with a speed of 180 km/h.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Angle of takeoff, θ = 30 degrees
- Speed of the airplane, v = 180 km/h
- Altitude to reach, h = 9 km

2. **Break Down the Velocity into Components:**
- The vertical component of the velocity (uy) can be calculated using the sine function:
\[ uy = v \cdot \sin(θ) = 180 \cdot \sin(30^\circ) \]
- Since \(\sin(30^\circ) = \frac{1}{2}\):
\[ uy = 180 \cdot \frac{1}{2} = 90 \text{ km/h} \]

3. **Calculate the Time to Reach the Altitude:**
- We need to find the time (t) it takes to reach an altitude of 9 km. We can use the formula:
\[ t = \frac{\text{Distance}}{\text{Speed}} = \frac{h}{uy} \]
- Substituting the values we have:
\[ t = \frac{9 \text{ km}}{90 \text{ km/h}} = \frac{1}{10} \text{ hours} \]

4. **Convert Time into Minutes:**
- To convert hours into minutes, we multiply by 60:
\[ t = \frac{1}{10} \cdot 60 = 6 \text{ minutes} \]

### Final Answer:
The time it will take for the airplane to reach an altitude of 9 km is **6 minutes**.

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