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