# Speed of sound

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A man dropped a rock into a well. He heard the sound of the rock hitting the water after 1.89 seconds. Considering the speed of sound to be 332 metres per second. How deep is the well?

This is a tricky question, since the rock has an acceleration when it's going down, and the sound have a constant velocity going back up.

You guys try it.

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The way i would go about doing this is using a simultaneous equation. But unfortunatly its not my homework question so im not going to figure it out.

Good luck

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Heh, it's not my homework. Just an interesting physics question for all of us that would requires abit of thinking.

Of course, it does involve a simultaneous equation, because the motion is divided into 2 parts since it have different time of flight.

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Distance up equals distance down. One distance equation will just be your kinematics equation d=.5aT^2, the other will be speed of sound (Vs) times the time it takes for the sound to get up the well (Ts). These two equations equal eachother, you will be able to solve for one of the T's in terms of another. You know that T+Ts=1.89. You now have two equations and two variables, solve how you like.

Not bad for a person who's high, although it did take me a few minutes to figure that out

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I got the answer of 65.8 metres. Did you got it too?

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There's an old B.C. cartoon that depcits this very scenario (the tagline is something like "Your well is 2 seconds deep") that was used on physics quizzes when I was teaching in the navy.

65.8 meters will take 3.66 seconds. Alway check your answer for reasonableness.

The answer has to be less than 17.5 meters. The large speed of sound is going to mean round-off will play a large role, so you have to carry extra digits in your intermediate calculations to avoid problems.

Your answer is about a factor of four to large. I get 16.6 m. 1.84 sec down, 0.05 up.

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swansont, your answer is right. After I posted the answer of 65.8 metres, I went to bed. While I was sleeping, I was thinking about this problem. I knew 65.8 metres was incorrect since the rock can't go down 65.8 metres in 1.89 seconds. Therefore, I woke up again, and redid my calculations. In the end, I got 16.6 metres too. It was too late (1:30 AM) for me to correct my post, so I'll do it tonight.

I'll explain this in more details later when I get home from school. I'll also post my calculations for the other people to see.

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S = (.5 * 9.8 * T1 * T1) = (340 * T2)

(4.9 * T1 * T1) = (340 * T2)

(T1 * T1) = (69.387 * T2)

(T1 * T1) / 69.387 = T2

Also T1 + T2 = 1.89

T1 + (T1 * T1 / 69.387) = 1.89

69.387T1 + (T1 * T1) = 131.14

Shuffle terms and drop the 1 from T1

T² + 69.387T - 131.34 = 0

T = (-69.387 +- sqrt(69.387² - 4 * -131.34)) / 2

T = (-69.387 +- sqrt(4814.55 + 525.36)) / 2

T = (-69.387 +- sqrt(5339.91)) / 2

T = (-69.387 +- 73.074) / 2

T = 3.7687 / 2

T = 1.84

T is T1

S = 4.9 * 1.84 * 1.84 * = 16.58m

Uh. I just couldn't let it go.

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Farsight, try to use LaTex next time, as it would be easier for us to read your calculations. I noticed that you did it in a little different way than I did.

My calculations:

As swansont stated that the distance equals both way, and that the rock have an acceleration due to gravity, you use $d=\frac{1}{2}at^2$. Also, for the sound, you use $d=vt$.

Keep in mind that $T_r$ (time for the rock), combined with $T_s$ (time for the sound) equals to $1.89$ seconds. ($T_r+T_s = 1.89s$).

1) $d=\frac{1}{2}(9.81 m/s^2)(1.89-T_s)^2$

$\rightarrow$Notice how $1.89-T_s$ leaves the time for $T_r$ (the amount of time for flight of the rock).

2) $d=(332 m/s)(1.89-T_r)$

Since the distance equals both way you have:

1 = 2

$\frac{1}{2}(9.81 m/s^2)(1.89-T_s)^2 = (332 m/s)(1.89-T_r)$

Now you have a quadratic equation:

$4.905(1.89-T_s)^2 + 332T_r - 627.48 = 0$

Using the quadratic formula you get: $1.839981808 s$ and another negative number which you don't need.

$T_s = 1.89 - T_r$

$T_s = 1.89 - 1.839981808 = 0.0500181917 seconds$ (That's the amount of time for the sound to arrive the human's ear at the top.

Using the distance formulas stated above, you input the 2 different times in their formula, and you will get the same distance for each.

The well is 16.60603963 metres deep.

I have another challenge physics problem for you guys and I'll post it tomorrow night.

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Sorry Evo, I was in a hurry.

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Here is another problem:

An applied force $F_{app}$ of 335 Newtons is required to accelerate a chair across a level floor at $0.722 m/s^2$. If the coefficient of kinetic friction is .330, find the mass of the chair.

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Anybody? I'm sure there is a lot of physicists out there!!

Tip: The answer that you might get is outrageous. (The chair is really HEAVY).

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Anybody? I'm sure there is a lot of physicists out there!!

I think you may be overestimating the satisfaction of, and time people are willing to commit to, working out a physics 101 problem.

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If I ignore friction and think about how many Newtons it would take to accelerate the chair to 1m/s/s we're talking about 335 / .722 = 464 Newtons and so 464kg. The coefficient of friction of .33 means the chair ain't as heavy as that, but to be honest, hands up, I don't know about friction and 464 * .67 = 311kg sounds too easy.

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The net force is given by:

$F_{app} - F_{friction} = F_{app} - \mu R$

where R is the reaction of the chair on the floor. We know the chair is on a level floor from the question, so R will be equal to mg (weight of the chair), therefore:

$\mu R = \mu m g$

Substituting in the numers give us:

$335 - 0.330*9.81*m$

Now by using F = ma

$335 - 0.330*9.81*m = ma$

and a = 0.722

doing a quick rearrangement (all the terms with mass in them moved to the right):

$335 = m(0.722 + 0.330*9.81)$

and then the final rearrangement to get an expression for mass.

$m = \frac{335}{0.722 + 0.330*9.81}$

therefore m = 84.6kg

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Thanks 5164. Of course, I forgot the weight of the chair. Duh.

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Good job 5614. See it was not too hard was it? I don't understand why many people won't participate in this. (Maybe they are not really that smart). But if you got a wrong answer, that doesn't matter, because you will learn from your mistakes.

The most general formula for forces is obviously $F_{net}=ma$.

I drew a FBD (Free Body Diagram), and you can see where your vectors' directions. So the $F_{net}$ is changed into:

$F_{app} - F_{fk} = ma$

$F_{app} = m(.722) + /mumg$

$335N = m(.722) + m(3.2373)$

$335N = m(3.9593)$

$m=84.6 kg$.

I'll post another physics question soon.

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I don't understand why many people won't participate in this.

Why would they? I usually have better things to do than solve high school physics problems.

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;309480'']Why would they? I usually have better things to do than solve high school physics problems.

OHHH!!!!!!!!!!! I was actually thinking that my problems were too hard for them to solve, so that was why people didn't participate.

I'll put up a question that is soo hard that you'll kick yourself all the night!!!

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The well is 16.60603963 metres deep.[/b]

Wrong. The well can be much deeper than the surface of its water.

That is not even the distance to the water level in the well.

The unknown size of the stone poses a much more complex calculation, as the larger it be in relation to the unknown well pipe diameter, causes more air cushioning its falling velocity.

The unknown gap between the stone and the well pipe wall can change calculations considerably.

Miguel

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Wrong. The well can be much deeper than the surface of its water.

That is not even the distance to the water level in the well.

The unknown size of the stone poses a much more complex calculation, as the larger it be in relation to the unknown well pipe diameter, causes more air cushioning its falling velocity.

The unknown gap between the stone and the well pipe wall can change calculations considerably.

Miguel

But this is a physics problem. In physics-world, there is no air resistance nor friction, unless specifically noted.

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Also, I stated in the question, that the rock hit the water, meaning the water surface, that's it. You have to read the question carefully.

My answer is right, I'm pretty sure.

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Wrong. The well can be much deeper than the surface of its water.

That is not even the distance to the water level in the well.

The unknown size of the stone poses a much more complex calculation, as the larger it be in relation to the unknown well pipe diameter, causes more air cushioning its falling velocity.

The unknown gap between the stone and the well pipe wall can change calculations considerably.

Miguel

Throw in any friction calculation and the resulting distance is shorter for the given time.

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Here is another physics problem. It's more challenge, and it's one of the questions from my university physics class. (Don't worry, I already passed my assignment in today). I have the answer for this question, so you can try the question yourself. If nobody can answer the question, then I'll give you hints.

PROBLEM: Consider the illustrated system. A constant force of 10N acts in the x direction on block B. What are the accelerations of blocks A and B? $m_A = 10 kg$, and $m_B = 15 kg$. Assume a frictionless interface between the two blocks and the ground.

(All the pulleys are actually the same size, my drawing just sucks).

Anybody?

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