Answer
$30\;cm$
Work Step by Step
In the part (a) of this problem, we have formulated that
$x=2\sqrt {h(H-h)}$
or, $h^2-Hh+\frac{x^2}{4}=0\;...............(1)$
The eq. 1 is a quadratic equation for $h$, when the values of $H$ and $x$ are fixed.
The solutions of $h$ are:
$h=\frac{H\pm\sqrt {H^2-x^2}}{2}$
Substituting the given values
$h=\frac{40\pm\sqrt {40^2-34.6^2}}{2}\;cm$
or, $h=\frac{40\pm20}{2}\;cm$
Thus the solutions are: $h_1=30\;cm$ and $h_2=10\;cm$
Therefore, a second hole should be made at a depth $30\;cm$ to give the same value of $x$
