A helicopter takes off along the vertical with an acceleration `a = 3m//s^(2)` and zero initial velocity. In a certain time the pilot switches off the engine. At the point of take off, the sound dies away in a time `t_(2) = 30 sec`. Determine the velocity of the helicopter at the moment when its engine is switched off assuming that velocity of sound is 320 m/s.

Answer

To solve the problem, we need to determine the velocity of the helicopter at the moment when its engine is switched off. The helicopter takes off with an acceleration of \( a = 3 \, \text{m/s}^2 \) and has an initial velocity of \( u = 0 \).

### Step 1: Determine the altitude \( h \) of the helicopter when the engine is switched off.

Using the equation of motion:
\[ h = u t_1 + \frac{1}{2} a t_1^2 \]
Since \( u = 0 \), the equation simplifies to:
\[ h = \frac{1}{2} a t_1^2 \]
Substituting \( a = 3 \, \text{m/s}^2 \):
\[ h = \frac{1}{2} \cdot 3 \cdot t_1^2 = \frac{3}{2} t_1^2 \]

### Step 2: Relate the time taken for sound to travel back to the ground.

The time taken for the sound to reach the ground after the engine is switched off is given as \( t_2 = 30 \, \text{s} \). The sound travels at a speed of \( c = 320 \, \text{m/s} \). The time taken for sound to travel distance \( h \) is given by:
\[ t_2 = \frac{h}{c} \]
Substituting the expression for \( h \):
\[ 30 = \frac{\frac{3}{2} t_1^2}{320} \]
Rearranging gives:
\[ 30 \cdot 320 = \frac{3}{2} t_1^2 \]
\[ 9600 = \frac{3}{2} t_1^2 \]
Multiplying both sides by \( \frac{2}{3} \):
\[ t_1^2 = \frac{9600 \cdot 2}{3} = 6400 \]
Taking the square root:
\[ t_1 = \sqrt{6400} = 80 \, \text{s} \]

### Step 3: Calculate the velocity \( v \) of the helicopter when the engine is switched off.

The velocity of the helicopter at the moment the engine is switched off can be calculated using:
\[ v = u + a t_1 \]
Since \( u = 0 \):
\[ v = a t_1 = 3 \cdot 80 = 240 \, \text{m/s} \]

### Step 4: Verify the calculations.

To ensure that the calculations are correct, we can check the values:
- The altitude \( h \) when the engine is switched off:
\[ h = \frac{3}{2} t_1^2 = \frac{3}{2} \cdot 6400 = 9600 \, \text{m} \]
- The time for sound to travel this distance:
\[ t_2 = \frac{h}{c} = \frac{9600}{320} = 30 \, \text{s} \]
This confirms that our calculations are consistent.

### Final Answer:
The velocity of the helicopter at the moment when its engine is switched off is:
\[ \boxed{240 \, \text{m/s}} \]