Take the stone's position at ground level to be the origin, and the downward direction to be negative. Then its position in the air [tex]y[/tex] at time [tex]t[/tex] is given by
[tex]y=100\,\mathrm m-\dfrac g2t^2[/tex]
Let [tex]d[/tex] be the depth of the well. The stone hits the bottom of the well after 5.00 s, so that
[tex]-d=100\,\mathrm m-\dfrac g2(5.00\,\mathrm s)^2\implies d=\boxed{22.6\,\mathrm m}[/tex]
