Kinematics in 1D problem

Oct 2013
48
0
Hi,

I have a problem like this:
A man drops a rock into a well.

(a) The man hears the sound of the splash 2.5 s after he releases the rock from rest. The speed of sound in air (at the ambient temperature) is 336 m/s. How far below the top of the well is the surface of the water?

Using the following transformations:

I. y_f = y_i +(v_yi)(t_1)+1/2(a_y)(t_1)^2

II. V_s(speed of sound)= y_f/t_2 => y_f = (V_s)(t_2)

III. t_1+t_2 = 2.5s => t_2 = 2.5s - t_1

Now, solving the system of equations and replacing t_2 with what we got in III:

-4.9t^2+336t_1 - 840=0 ,

solving for t_1 as an equation of second degree I get t_1 = 2.5984,

I am a bit confused, since I should have got something lesser then 2.5s,

but anyways, taking the difference between these two times and subtracting it from 2.5 s I get t_1 = 2.365

and using this time in equation I. I get y_f = 28.260m, which is correct according to my answer in the book...


Now, I have to answer to the second point:

(b) If the travel time for the sound is ignored, what percentage error is introduced when the depth of the well is calculated (in %)?

considering t_1 = t = 2.5s,

we get y_f = 30.625m, which is (30.625m - 28.260m) = 2.365m more then the real depth.

and finally, (2.365m)*(100%)/(28.260m) = 8.3687% , which is not considered as a correct answer; it says its more then 10% from the value of the correct answer.


I have two questions:

1. Why I get that confusing result when I solve for t_1?

2. What's my mistake when finding the percent of error?

Thanks
 

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
For the first problem you have a simple math error with the sign of one of the terms: the quadratic equation you need to solve is:

4.9t^2 +336t-840=0

This yields t=2.41 sec and depth of 28.6m, not 28.26m

For the second, using the value for depth that you got for (a) does indeed give an error of 8.3%, as you calculated. Using the correct value for the depth of 28.6m, I get an error 2.05m, and the error percent is 2.05/28.6 = 7.2%.
 
Oct 2013
48
0
For the first problem you have a simple math error with the sign of one of the terms: the quadratic equation you need to solve is:

4.9t^2 +336t-840=0

This yields t=2.41 sec and depth of 28.6m, not 28.26m

For the second, using the value for depth that you got for (a) does indeed give an error of 8.3%, as you calculated. Using the correct value for the depth of 28.6m, I get an error 2.05m, and the error percent is 2.05/28.6 = 7.2%.
thanks,

now I see my mistakes, still it wasn't a math error, I considered the acceleration of a free falling body with (-) sign, but it should be positive since it's in the same direction with body's movement... if I get it correctly
 
Last edited:

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
As long as you're consistent you can do it that way. For example you can say that the net displacement of the object falling (negative direction) plus the sound rising (in a positive direction) is 0. This yields:

-4.9t^2 + 336(2.5-t) = 0

Or you can say that the magnitude of the distance that the object falls is the same as the magnitude of the distance traveled by sound:

4.9t^2 = 336(2.5-t)

Either approach gives the same result.
 
Last edited:
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Oct 2013
48
0
As long as you're consistent you can do it that way. For example you can say that the net displacement of the object falling (negative direction) plus the sound rising (in a positive direction) is 0. This yields:

-4.9t^2 + 336(2.5-t) = 0

Or you can say that the magnitude of the distance that the object falls is the same as the magnitude of the distance traveled by sound:

4.9t^2 = 336(2.5-t)

Either approach gives the same result.
I see, thank you a lot