Quick speed of sound question?

[whatwhat11]

A swimmer sees a parachutist hit the water and hears the impact twice, once through the water and the second time through the air, 1.0 s later. How far from the swimmer did the impact occur?
Vs air = 340 m/s Vs water = 1400 m/s the answer is 450m

I don't understand what to do

 

[Chestermiller]

A swimmer sees a parachutist hit the water and hears the impact twice, once through the water and the second time through the air, 1.0 s later. How far from the swimmer did the impact occur?
Vs air = 340 m/s Vs water = 1400 m/s the answer is 450m

I don't understand what to do

This is a straightforward rate x time = distance (r t = d) problem.

Let d = distance from swimmer to splash point.

In terms of d, how much time does the sound take to get from the splash point to the swimmer in air?

In terms of d, how much time does the sound take to get from the splash point to the swimmer in water?

In terms of d, what is the difference between these two times?

(Actually, the answer is not 450m, it's only 449m, but who's counting?)

 

[whatwhat11]

This is a straightforward rate x time = distance (r t = d) problem.

Let d = distance from swimmer to splash point.

In terms of d, how much time does the sound take to get from the splash point to the swimmer in air?

In terms of d, how much time does the sound take to get from the splash point to the swimmer in water?

In terms of d, what is the difference between these two times?

(Actually, the answer is not 450m, it's only 449m, but who's counting?)

Vsound in air= (340)(2)
=680m

V sound in water = (1400)(1)
=1400

1400-680
doesn't equal 450m or 459m

 

[Chestermiller]

Vsound in air= (340)(2)
=680m

V sound in water = (1400)(1)
=1400

1400-680
doesn't equal 450m or 459m

Hu?

The answer to my first question was: the amount of time it takes for the sound to travel through the air from the splash to the swimmer is d/340 seconds

The answer to my second question was: the amount of time it takes for the sound to travel through the water from the splash to the swimmer is d/1400 seconds

The answer to my third question was: the difference between these two times is (d/340) - (d/1400) =1 second

Are you able to solve this equation for the distance d? If so, what do you get for d?

 

[Nickie]

I can help explain the process of determining the distance of the impact in this scenario. The key factor here is the difference in speed of sound between air and water. Sound travels much faster in water (1400 m/s) compared to air (340 m/s). This is because the density of water is much higher than air, allowing sound waves to travel more quickly.

Now, let's break down the information given in the question. The swimmer hears the impact twice - once through the water and the second time through the air. This means that the sound waves from the impact took two different paths to reach the swimmer's ears. The first path was through the water, which took less time (1.0 s) compared to the second path through the air.

To determine the distance of the impact, we can use the formula distance = speed x time. In this case, we need to calculate the distance for both paths separately and then add them together to get the total distance.

For the first path through the water, we know the speed of sound in water is 1400 m/s and the time taken is 1.0 s. Plugging these values into the formula, we get 1400 m/s x 1.0 s = 1400 m. This is the distance the sound waves traveled through the water.

For the second path through the air, we know the speed of sound in air is 340 m/s and the time taken is 1.0 s. Using the formula again, we get 340 m/s x 1.0 s = 340 m. This is the distance the sound waves traveled through the air.

Now, to get the total distance, we simply add these two distances together: 1400 m + 340 m = 1740 m. This is the total distance the sound waves traveled to reach the swimmer's ears. However, this includes the distance traveled through the water, which we don't need. So, we subtract the distance traveled through the water (1400 m) from the total distance to get the distance of the impact: 1740 m - 1400 m = 340 m.

Therefore, the impact occurred 340 m away from the swimmer. I hope this explanation helps clarify the process for determining the distance in this scenario.