Solve for Echo Time in Saltwater and Air - Physics Problem

[bilalsyed25]

The Question:
A boat is floating at rest in dense fog near a large cliff. The captain sounds a horn at water level and the sound travels through the salt water (1470 m/s) and the air (340 m/s) simultaneously. The echo in the water takes 0.40s to return. How much additional time will it take the echo in the air to return?

relevant equations: v=d/t

My answer:
d = v∆t

d = 1470*0.40

d = 588/2 (I checked online for this question, some ppl got different answers due to this part; 558/2 or just 588?)

d = 294m

294/340 = 0.86s

0.86-0.40 = 0.46

Additional time = 0.46s

 

[Doc Al]

d = 588/2 (I checked online for this question, some ppl got different answers due to this part; 558/2 or just 588?)

d = 294m

Realize that the sound must travel to the cliff and back.

 

[PeroK]

The Question:
A boat is floating at rest in dense fog near a large cliff. The captain sounds a horn at water level and the sound travels through the salt water (1470 m/s) and the air (340 m/s) simultaneously. The echo in the water takes 0.40s to return. How much additional time will it take the echo in the air to return?

relevant equations: v=d/t

My answer:
d = v∆t

d = 1470*0.40

d = 588/2 (I checked online for this question, some ppl got different answers due to this part; 558/2 or just 588?)

d = 294m

294/340 = 0.86s

0.86-0.40 = 0.46

Additional time = 0.46s

If something takes $0.4s$ to cover a distance and something else takes an extra $0.46s$ to cover the same distance, then the faster thing must be moving approx twice as fast as the slower thing. Is $1,470m/s$ approx twice $340 m/s$?

Using this logic can you estimate the answer (approx) in your head?

 

[Nestory]

Okay, The distance from water is 294 . But for time for echo via air to be heard , t: (2*294)/340 ; 1.73 so addirional time is 1.73-0.4; 1.3seconds

 

[Nestory]

The Question:
A boat is floating at rest in dense fog near a large cliff. The captain sounds a horn at water level and the sound travels through the salt water (1470 m/s) and the air (340 m/s) simultaneously. The echo in the water takes 0.40s to return. How much additional time will it take the echo in the air to return?

relevant equations: v=d/t

My answer:
d = v∆t

d = 1470*0.40

d = 588/2 (I checked online for this question, some ppl got different answers due to this part; 558/2 or just 588?)

d = 294m

294/340 = 0.86s

0.86-0.40 = 0.46

Additional time = 0.46s

Why you didnt multiply the distance 294m when finding the time through the air?

 

[PeroK]

Why you didnt multiply the distance 294m when finding the time through the air?

This thread is three years old and the OP has hopefully graduated by now.