Answer
The well has a depth of $~~40.7~m$
Work Step by Step
Let $h$ be the depth of the wall.
Let $t_1$ be the time the stone drops to the bottom.
Let $t_2$ be the time that the sound travels from the bottom of the well to the ear.
Note that $t_1+t_2 = 3.00~s$
We can use the time the stone drops to the bottom to find an expression for the height:
$h = \frac{1}{2}gt_1^2$
We can use the time the sound travels from the bottom of the well to the ear to find an expression for the height:
$h = v~t_2$
$h = v~(3.00-t_1)$
We can equate the two expressions to find $t_1$:
$h = \frac{1}{2}gt_1^2 = v~(3.00-t_1)$
$4.9t_1^2+(343)t_1-(343)(3.00) = 0$
$4.9t_1^2+(343)t_1-1029 = 0$
We can use the quadratic formula to find $t_1$:
$t_1 = \frac{-343 \pm \sqrt{(343)^2-(4)(4.9)(-1029)}}{(2)(4.9)}$
$t_1 = \frac{-343 \pm 371.24}{9.8}$
$t_1 = -72.9~s, 2.881~s$
We can choose the positive solution.
We can find $h$:
$h = \frac{1}{2}gt_1^2$
$h = \frac{1}{2}(9.80~m/s^2)(2.881~s)^2$
$h = 40.7~m$
The well has a depth of $~~40.7~m$
