Calculating Percentage Error in Canyon Depth Ignoring Sound Time

[munther]

A stone is dropped into a deep canyon and is heard to strike the bottom 13.4 s after release (the speed sound is 343m/s) What would be the percentage error in the depth if the time required for the sound to reach the canyon rim were ignored?

I found the depth of the canyon =2298m
then what is the next step?

 

[Doc Al]

How did you do calculate the depth?

The point is to calculate it two ways: once ignoring the time for the sound to travel; once taking it into account.

 

[munther]

sorrry
the distance = 1/2 a t^2
= .5 x 9.8 x (13.4/2)^2 = 219.961 m^2


that is when i take the time into account

but how i will not take it??

 

[Doc Al]

the distance = 1/2 a t^2
= .5 x 9.8 x (13.4/2)^2 = 219.961 m^2

The time needed here is the time that the stone was falling. Why did you divide by 2?

Think of the time to hear the echo as having two parts:
(1) The time that the stone was falling.
(2) The time it takes for the sound of the stone hitting the canyon floor to travel up to the top.

Time #1 + Time #2 = 13.4 seconds

Since sound travels pretty fast compared to the stone, time #2 will be small.

 

[munther]

ok that is very good

then how i will divide the time between the falling and the echo

 

[Doc Al]

then how i will divide the time between the falling and the echo

To incorporate the time delay due to the speed of sound, you'll need to set up equations and solve for the times. Hint: Set up two equations (one for the falling rock; one for the sound) relating distance and time.

 

[munther]

ok

d= .5xgxt1 = 343t2
t1+t2 =13.4
by solving both equations
t1= 11.51
t2= 1.89

then the depth = 648.9 m

is it right


next step:

%=(depth1-depth2)x100/(depth1)

we found one depth

the second depth when we ignored the time
is the second depth when time = 13.4 for the d=.5x9.8xt

 

[Doc Al]

ok

d= .5xgxt1 = 343t2
t1+t2 =13.4
by solving both equations
t1= 11.51
t2= 1.89

then the depth = 648.9 m

is it right

Looks good to me.

is the second depth when time = 13.4 for the d=.5x9.8xt

Right. (You have a typo in your formula--that should be t^2.)

 

[munther]

thanks

i posted it ...but it is wrong!

(879.844-648.9)/(879.844)=26.2% xxx Wrong xxx !

why??

 

[Doc Al]

thanks

i posted it ...but it is wrong!

(879.844-648.9)/(879.844)=26.2% xxx Wrong xxx !

why??

You need to read the question very carefully. I think they are asking for the percentage error the incorrect method (that ignores the sound travel time) introduces with respect to the correct method. The correct measurement should be in the denominator.