Solve for Echo Time in Saltwater and Air - Physics Problem

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The discussion revolves around calculating the additional time it takes for an echo to return in air after a sound travels through saltwater. The captain's horn sounds, and the echo in water takes 0.40 seconds to return, leading to a calculated distance of 294 meters. The time for the echo to return in air is then calculated as 0.86 seconds, resulting in an additional time of 0.46 seconds compared to the water echo. Some participants debate the calculations and mention discrepancies found online regarding the distance used. Ultimately, the conversation highlights the importance of accurate distance measurement and time calculation in physics problems.
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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?

Relevent 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
 
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bilalsyed25 said:
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.
 
bilalsyed25 said:
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?

Relevent 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?
 
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
 
bilalsyed25 said:
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?

Relevent 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?
 
Nestory said:
Why you didnt multiply the distance 294m when finding the time through the air?
:welcome:

This thread is three years old and the OP has hopefully graduated by now.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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