Answer:
Approximately [tex]19.6\; {\rm m}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] and that air resistance is negligible.)
Explanation:
Assume that the air resistance on the rock is negligible. During the descent, the acceleration [tex]a[/tex] of the rock will be constant: [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex].
It is given that the descent took [tex]t = 2.00\; {\rm s}[/tex]. Let [tex]x[/tex] denote the displacement (change in position) of the stone. Apply the SUVAT equation [tex]x = (1/2)\, a\, t^{2}[/tex] to find displacement [tex]x\![/tex]:
[tex]\begin{aligned}x &= \frac{1}{2}\, a\, t^{2} \\ &= \frac{1}{2}\, ((-9.81)\; {\rm m\cdot s^{-2}})\, (2.00\; {\rm s})^{2} \\ &\approx (-19.6)\; {\rm m}\end{aligned}[/tex].
Note that [tex]x[/tex] is negative since the water is below the initial position of the rock. Therefore, the distance to the water will be approximately [tex]19.6\; {\rm m}[/tex].