# Homework Help: Speed of sound of stone

1. Nov 16, 2007

### sun

1. The problem statement, all variables and given/known data

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.

2. Relevant 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)

3. 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

2. Nov 16, 2007

### 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.

3. Nov 16, 2007

### mgb_phys

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

4. Nov 18, 2007

### sun

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?

5. Nov 18, 2007

### hage567

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.

6. Nov 18, 2007

### 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 :)

Last edited: Nov 18, 2007
7. Nov 18, 2007

### 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.

8. Nov 18, 2007

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?

9. Nov 18, 2007

### Dick

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.

10. Nov 18, 2007

### 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

11. Nov 18, 2007

### 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.

12. Nov 18, 2007

### 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?

13. Nov 18, 2007

### Dick

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.