Let's break down the problem and solve it step by step.
1. **First scenario**:
- The siren is sounded and the echo is heard after 12 seconds.
- Let's denote the distance between the ship and the cliff as \( d \).
- The time it takes for the sound to travel from the ship to the cliff and back to the ship (echo) is 12 seconds.
2. **Second scenario**:
- The siren is sounded again, and the echo is heard 8 seconds later.
- The time it takes for the sound to travel from the ship to the cliff and back to the ship (echo) is 8 seconds.
Given the speed of sound in air is 340 m/s, we can use the formula:
\[
\text{Distance} = \text{Speed} \times \text{Time}
\]
Let's denote the velocity of the ship as \( v \).
1. In the first scenario:
- The time it takes for the sound to travel to the cliff and back is 12 seconds.
- So, the distance traveled by sound is \( 2d \).
- Using the formula, we get:
\[
2d = 340 \times 12
\]
\[
d = \frac{340 \times 12}{2} = 2040 \text{ meters}
\]
2. In the second scenario:
- The time it takes for the sound to travel to the cliff and back is 8 seconds.
- So, the distance traveled by sound is again \( 2d \).
- Using the formula, we get:
\[
2d = 340 \times 8
\]
\[
d = \frac{340 \times 8}{2} = 1360 \text{ meters}
\]
Now, we need to find out the velocity at which the ship was approaching the cliff.
The difference between the distances covered by the ship in the two scenarios is the distance it covers in 2.1 minutes (or 126 seconds).
The ship moves \( 2040 - 1360 = 680 \) meters in 126 seconds.
So, the velocity of the ship is:
\[
v = \frac{680 \text{ meters}}{126 \text{ seconds}} \approx 5.396 \text{ m/s}
\]
Therefore, the velocity at which the ship was approaching the cliff is approximately \( 5.396 \) m/s.
