When speed of sound is not negligible

[OpticalSuit]

Homework Statement

A stone is dropped from rest at the opening of a well, 2.40s later a splash is heard. how deep is the well?
given that speed of sound is 336m/s.

Homework Equations

x_f=x_i+v_it+1/2at²

The Attempt at a Solution

What does one do to account for the time it takes for the sound to bounce back?!?

 

[willem2]

Homework Statement

A stone is dropped from rest at the opening of a well, 2.40s later a splash is heard. how deep is the well?
given that speed of sound is 336m/s.

Homework Equations

x_f=x_i+v_it+1/2at²

The Attempt at a Solution

What does one do to account for the time it takes for the sound to bounce back?!?

If the well has a depth d, what is the time for the sound to come out again (as a function of d)

Add this to the time the rock takes to fall to this depth, and the outcome should be 2.4s
This will allow you to find d.

 

[OpticalSuit]

thanx for the reply but i figured out the problem in the end
here is what i wrote:
The sound is heard after 2.40 seconds and sound travels 336m/s. And we know the acceleration due to gravity is -9.8m/s^2. The distance the rock travels is equal to the distance sound traveled x1=x2. And the time the rock took to hit the water added to the time the sound came up is equal to 2.40 s t1+t2=2.40s.
We must solve for one of the times in order to find the distance.
The distance equation for the rock is x1=1/2gt^2 since the intial velocity and distance is 0. For sound we have x2=vt. Therefor 1/2gt1^2=vt2, now we bring vt2 over to get (1/2)gt1^2-vt2=0. Since we know t1+t2=2.40s, therefore t2=2.40s-t1. Plugging that in gives us ½(-9.8)t1^2-(336)(2.40s-t1)=0 which gives us ½(-9.8)t1^2+336t1-806.4. now using the quadratic formula we can solve for t1 which gives us two values of either -2.32 or 70.9. logically we will take 2.32seconds for the rock to hit the water. Plugging that into the original distance equation of the rock gives us ½(9.8)(2.32)^2=-26.4 meters