Answer:
[tex](-19.6)\; {\rm m}[/tex], assuming that the object was initially not moving.
Explanation:
Under the assumptions, the acceleration [tex]a[/tex] of this object will be constant: [tex]a = g = (-9.8)\; {\rm m\cdot s^{-2}}[/tex].
The initial velocity [tex]u[/tex] of this object will be [tex]0\; {\rm m\cdot s^{-1}}[/tex] since this object was initially not moving: [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex].
Let [tex]x[/tex] denote the displacement of this object. This object has been accelerating for [tex]t = 2.0\; {\rm s}[/tex]. Apply the SUVAT equation [tex]x = (1/2)\, a\, t^{2} + u\, t[/tex] to find the displacement of this object in that much time:
[tex]\begin{aligned}x &= \frac{1}{2}\, a\, t^{2} + u\, t \\ &= \frac{1}{2} \times (-9.8)\; {\rm m\cdot s^{-2}} \times (2.0\; {\rm s})^{2} + 0\; {\rm m\cdot s^{-1}} \times 2.0\; {\rm s} \\ &= \frac{1}{2} \times (-9.8) \times 2.0^{2} \; {\rm m} \\ &= (-19.6)\; {\rm m}\end{aligned}[/tex].
In other words, the displacement of this object would be [tex](-19.6)\; {\rm m}[/tex] (displacement is negative since this object is now below where it initially was.)