Let the body start from rest and fall vertically downward.
Suppose the upper point $A$ is at a distance $x$ from the starting point.

Velocity at point $A$
Using $v_A^2 = 2gx$ with $g=10\,\text{m s}^{-2}$, $v_A^2 = 20x$ so $v_A = \sqrt{20x}$

Motion from $A$ to $B$
The body travels from $A$ to $B$ in $t=2\,\text{s}$, covering distance $s=80\,\text{m}$ under acceleration $g=10\,\text{m s}^{-2}$.

Using $s = v_A t + \frac{1}{2}gt^2$ $80 = v_A(2) + \frac{1}{2}(10)(2^2)$ $80 = 2v_A + 20$ $2v_A = 60$ $v_A = 30\,\text{m s}^{-1}$

Now find $x$
From $v_A^2 = 20x$ $30^2 = 20x$ $900 = 20x$ $x = 45\,\text{m}$

Final Answer
The distance of point $A$ from the starting point is $\boxed{45\,\text{m}}$

Comparison with stored answer
Stored correct answer = $45$
Derived answer = $45$
Hence, they agree.