# Need help with this sound problem

1. Homework Statement

A rock was dropped into a deep canyon and you hear the rock hit the bottom after 11.3 seconds.How deep is the canyon?
t=11.2s
velocity of sound is 340m/s in air
gravity is 9.8m/s/s

2. Homework Equations

i know that only two equations are involved in this problem, they are
d=vt
d=Vit+1/2at^2 (Vi is initial velocity, since Vi equal to 0 the equation is just d=1/2at^20)

3. The Attempt at a Solution
this is what i did:
(340m/s x 11.3s)/2=1921m
by the way i divided by 2 because the rock need 5.65( 11.3/2) to reach the bottome and another 5.65 seconds for the sound to reach your ears. Total of 11.3 seconds

thank you a lot

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dynamicsolo
Homework Helper
You're right that there are two terms to look at. But the speed of sound is a lot faster than that rock is ever going to move, so the sound can't have been traveling for 11.3 seconds...

There are two parts to this problem to put together. You don't know the depth of the canyon, but you can figure out how long a falling rock, starting from rest, will drop in time t. So you have a relation between the depth of the canyon, D, and the time, t, it will take the rock to hit bottom.

Once the rock hits bottom, the impact will now produce a sound which will travel, among other directions, back up through that depth D to reach your ear. If we call that time t' , how is that related to D?

The only thing you know directly is that the falling time t and the sound-travel time t' add up to 11.3 seconds. Since you've written both of these in terms of D, you have just one equation in one unknown to solve in order to find D.

i know what you saying but when i tried to find t of the falling rock, i got 19s which doeasnt make any sense because like you said: the falling time and the sound travel time must add up to 11.3 ,
here is what i did: since the equation for distance is D=1/2at^2
so d=1921m and a=9.8m/s^2
plugged all number in, solved for t and i got 19s
i know that a big portion of 11.3 seconds is time for the rock to hit the bottom and just a little of 11.3s for the sound to reach your ear?
what did i do wrong?
if 1921 is not D what is 1921?
thank for helping me

dynamicsolo
Homework Helper
i know what you saying but when i tried to find t of the falling rock, i got 19s which doeasnt make any sense because like you said: the falling time and the sound travel time must add up to 11.3 ,
here is what i did: since the equation for distance is D=1/2at^2
so d=1921m and a=9.8m/s^2
plugged all number in, solved for t and i got 19s
I'm saying that you can't get a number yet: you don't know the time the rock takes to fall, so you can't find D yet.

You have D = (1/2)g(t^2) , so

(t^2) = 2D/g , or

t = sqrt(2D/g) .

This is the time during which the rock falls.

The time the sound of the impact will take to return is

t' = D/v_sound .

It is this total time that is 11.3 seconds.

So you need to solve

sqrt(2D/9.81) + (D/340) = 11.3 .

I don't know what 1921 is... Even if the rock falls for 11.3 seconds, the distance covered will only be 626 m. (This will actually be in the general neighborhood of the right answer, but since it would take 1.8 seconds for the sound to return from this distance, the canyon must be less deep than that. I would estimate the answer will be somewhat less than 500 meters...)

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but 1921m mean nothing in this problem?

1921m means nothing because you hear the sound of the rock falling 11.3 seconds after you let it go, not 11.3 seconds after the rock hits the bottom of the canyon.

dynamicsolo
Homework Helper
but 1921m mean nothing in this problem?
How are you getting a value of 1921? What are you calculating?

dynamicsolo
Homework Helper
1921m means nothing because you hear the sound of the rock falling 11.3 seconds after you let it go, not 11.3 seconds after the rock hits the bottom of the canyon.
Aha, now I see where 1921 comes from. But the impact does not happen halfway between the release of the rock and the time you hear the impact. So the depth is definitely not one-half of (11.3 seconds)·(340 meters/second). (The sound takes only a little over 10% of the total interval to travel, so the impact happens almost 90% of the way from the release to the return of the sound...)