How Do You Calculate the Depth of a Well Using the Speed of Sound?

[sun]

Homework Statement

A stone is dropped from rest into a well. The sound of the splash is heard exactly 1.50 s later. Find the depth of the well if the air temperature is 6.0°C.

Homework Equations

i'm not sure what equation i need to use here.
i know that 6C=279.15K
v=dx/dt
v=(331m/s)*sqrt(Temp/273K)

The Attempt at a Solution

I'm a bit confused as to how to start this problem. Any suggestions would be much appreciated.

Thank you

 

[Dick]

You also need a kinematical equation to figure out how long it takes the stone to drop to the water surface. Something like y(t)=y0-(1/2)*g*t^2.

 

[mgb_phys]

Your equation for speed of sound is slightly wrong
c = 331 * sqrt( 1 + T/273) T in degrees C

 

[sun]

You also need a kinematical equation to figure out how long it takes the stone to drop to the water surface. Something like y(t)=y0-(1/2)*g*t^2.

Doesn't the problem already state that it takes 1.5s for the stone to reach the surface of the water?

I've determined that the speed of sound in air at 6C is 334.62m/s.

Using a kinematic equation i know that at dt=1.5s and Vo=0 then dy is 11.03m
dy=Vo(t)+.5(9.8)(t)^2

But how does the different media that the sound travels through play a role in this?

 

[hage567]

Doesn't the problem already state that it takes 1.5s for the stone to reach the surface of the water?

No, it tells you that the sound of the splash is heard 1.5 seconds later. That time includes the time for the stone to fall to the water and the time for the sound wave to travel back up the cliff.

 

[sun]

So,I know that it takes 1.5s for the stone to free fall down to the water. Also, the sound traveling back up travels at a speed of 334.62m/s.

If i figure out how long it takes the sound to travel back up the well, i can subtract that number from 1.5s. Then i will be able to solve the original question.

Although I'm still unsure how to solve the problem am I on the correct path? Suggestions would be much appreciated.

thank you :)

 

[Dick]

No. It doesn't take 1.5sec to fall to the water. Time to fall is a function of the depth of the well, d. So is the time for the sound to go back up. Work out those two functions of the unknown d, and set their sum=1.5sec. Then solve for d.

 

[Izmad]

y(t)=y0-(1/2)*g*t^2

y(t)=0-(1/2)*9.8*1.5^2 = 11.03m

So, the distance to travel down and back up would be 22.06m?

 

[Dick]

y(t)=y0-(1/2)*g*t^2

y(t)=0-(1/2)*9.8*1.5^2 = 11.03m

So, the distance to travel down and back up would be 22.06m?

You are making the same mistake sun is. Time to fall isn't 1.5sec. The round trip time is 1.5 sec. Just leave t an unknown. Solve for it in terms of y. Now add that to the expression for t in terms of y for the sound. The sum is 1.5sec. Now solve for y.

 

[sun]

this is what i did.

v=dx/dt, so 334.62m/s=11.03/dt. dt=.033s. This is the time it took the sound to travel from the water back up the well.

I plugged .033s into y(t)=y0-(1/2)*g*t^2 and got 10.54m. Which supposedly is the depth of the well.

thanks for the patience and help.
It is much appreciated

 

[Dick]

If the depth of the well is 10.54m, then what is 11.03m? You aren't being consistent, you're using two different depths. The time down is t1=sqrt(2y/g). The time up is t2=y/vsound. t1+t2=1.5sec.

 

[hiddenbyleaves]

Hi Dick,
I got:

sqrt(2y/g)+9y/339.47)=2.4 but can't solve the equation. I keep on getting weird and very big numbers. Can you help?

 

[Dick]

Hi Dick,
I got:

sqrt(2y/g)+9y/339.47)=2.4 but can't solve the equation. I keep on getting weird and very big numbers. Can you help?

I don't have a clue where you are getting those numbers. Also don't know why I'm getting so many posts from people besides the original poster. But to solve something like that just get the sqrt part by itself. So you have sqrt(2y/g)=a+by. You figure out what a and b are. Now square both sides. Now you have a quadratic equation. There is a formula for the roots.