When sound is produced in an aeroplane with a velocity of 200 m/s horizontally its echo is heard after `10 sqrt5` seconds. If velocity of sound in air is `300"m s"^(-1)` the elevation of aircraft is

Options

A. 250 m
B. `250 sqrt(5)` m
C. 1250 m
D. 2500 m

Explanation

To find the elevation of the aircraft, we can follow these steps:

### Step 1: Understand the problem
We have an airplane producing sound while moving horizontally. The echo of the sound is heard after a certain time. We need to find the height of the aircraft above the ground.

### Step 2: Identify the given values
- Velocity of sound in air (Vs) = 300 m/s
- Velocity of the airplane (Vt) = 200 m/s
- Time taken to hear the echo (t) = \(10\sqrt{5}\) seconds

### Step 3: Calculate the total distance traveled by sound
The total distance traveled by the sound in time \(t\) is given by:
\[ \text{Total distance} = Vs \times t \]
Substituting the values:
\[ \text{Total distance} = 300 \, \text{m/s} \times 10\sqrt{5} \, \text{s} = 3000\sqrt{5} \, \text{m} \]

### Step 4: Determine the distance to the ground (BP)
Since the sound travels to the ground and back, the distance from the aircraft to the ground (BP) is half of the total distance:
\[ BP = \frac{\text{Total distance}}{2} = \frac{3000\sqrt{5}}{2} = 1500\sqrt{5} \, \text{m} \]

### Step 5: Calculate the distance traveled by the airplane (AB)
The distance traveled by the airplane in time \(t\) is:
\[ AB = Vt \times t = 200 \, \text{m/s} \times 10\sqrt{5} \, \text{s} = 2000\sqrt{5} \, \text{m} \]

### Step 6: Use the Pythagorean theorem
We can form a right triangle where:
- BP is the height (PC) we want to find,
- AB is the horizontal distance traveled by the airplane,
- AP is the distance from the point where the sound was produced to the point where it hit the ground.

Using the Pythagorean theorem:
\[ AB^2 = BP^2 + PC^2 \]
Substituting the values:
\[ (2000\sqrt{5})^2 = (1500\sqrt{5})^2 + PC^2 \]

### Step 7: Solve for PC
Calculating the squares:
\[ (2000\sqrt{5})^2 = 4000000 \cdot 5 = 20000000 \]
\[ (1500\sqrt{5})^2 = 2250000 \cdot 5 = 11250000 \]
Now substituting back into the equation:
\[ 20000000 = 11250000 + PC^2 \]
Rearranging gives:
\[ PC^2 = 20000000 - 11250000 = 8750000 \]
Taking the square root:
\[ PC = \sqrt{8750000} = 2958.3 \, \text{m} \]

### Conclusion
The elevation of the aircraft is approximately \(2958.3 \, \text{m}\).