Simple problem. Depth of well providing sound reaches you in 1.5 second.

In summary, the problem at hand is to find the depth of a well by dropping a rock and timing the sound of the splash. The equations used are t1+t2=1.5s, x=.5*a*t^2, and v=x/t, with a known speed of sound of 343 m/s. By substituting the equations and solving with the quadratic formula, the depth can be determined. Another approach is to use the equation 343*t2=4.9*(t2-1.5)^2, which can also be solved using the quadratic formula.
  • #1
new324
30
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I was wondering you someone could help me with an alternative way of solving this problem. Just to help my algebra a bit.
You drop a rock down a well. Find the depth of the well providing the sound of the splash reaches you in 1.5 seconds.

First off, we know speed of sound is 343 m/s.
Here are my equations:
t1+t2=1.5 s t1 is time for rock to hit bottom t2 is time for sound to reach your ears.
x=.5*a*t^2 -> x=4.9*t1^2
v=x/t -> x=343*t2

Now I sub the last 2 equations to get:
343*t2=4.9*t1^2
Now we sub. for t2:
343*(1.5-t1)=4.9*t1^2
From here we can reduce, and solve with the quadratic formula.

I was wondering how you would solve the equation, if you sub. for t1 instead of t2. This would make the equation:
343*t2=4.9*(t2-1.5)^2
 
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  • #2
new324 said:
I was wondering how you would solve the equation, if you sub. for t1 instead of t2. This would make the equation:
343*t2=4.9*(t2-1.5)^2
It's just another quadratic equation. Multiply out the term on the right (do the square) and put it into standard form.
 
  • #3


Thank you for sharing your method for solving this problem. Here is an alternative approach using algebra:

Let d be the depth of the well and t be the time it takes for the sound to reach you. We can use the formula d = rt, where r is the speed of sound, to find the depth.

Since we know the speed of sound is 343 m/s, we can write the equation as d = 343t.

Now, we also know that the total time, t, is equal to the sum of the time it takes for the rock to hit the bottom of the well and the time it takes for the sound to reach you. So we can write the equation as t = t1 + t2.

Since we are given that t = 1.5 seconds and t1 is the time it takes for the rock to hit the bottom, we can write the equation as 1.5 = t1 + t2.

Now, we can substitute this value of t into our first equation to get d = 343(1.5 - t1).

Using the equation x = 4.9t^2, we can also write t1 as √(d/4.9).

Substituting this value into our equation 1.5 = t1 + t2, we get 1.5 = √(d/4.9) + t2.

Solving for t2, we get t2 = 1.5 - √(d/4.9).

Now, we can substitute this value of t2 into our first equation to get d = 343(1.5 - √(d/4.9)).

Simplifying this equation, we get d = 514.5 - 343√(d/4.9).

Solving for d, we get a quadratic equation: 343√(d/4.9) + d = 514.5.

Solving this equation using the quadratic formula, we get two possible solutions: d = 0.45 m or d = 173.68 m.

So, using this method, we can find that the depth of the well is either 0.45 m or 173.68 m.
 

1. How can the depth of a well be determined using sound?

The speed of sound in air is approximately 343 meters per second. By dividing the time it takes for sound to travel from the bottom of the well to the surface (1.5 seconds) by the speed of sound, we can determine that the depth of the well is approximately 514.5 meters.

2. Is the speed of sound constant in all media?

No, the speed of sound can vary depending on the medium it is traveling through. For example, sound travels faster in solids than in liquids or gases.

3. How accurate is the calculated depth of the well using sound?

The calculated depth of the well using sound is fairly accurate, but it may be affected by external factors such as air temperature and humidity. The speed of sound can also vary slightly depending on the composition of the well and the surrounding soil or rock.

4. Can sound be used to determine the depth of other structures or objects?

Yes, sound can be used to determine the depth of other structures or objects as long as the speed of sound in the medium is known. This method is commonly used in geology and oceanography to measure the depth of the ocean floor or underground layers of rock.

5. Are there any limitations to using sound to determine the depth of a well?

While sound can provide a fairly accurate estimate of the depth of a well, it may not be suitable for deep wells or wells with complex structures. Additionally, external factors such as echoes or interference can affect the accuracy of the measurement.

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