a
\(v =1.25 \times 8 m / s =10\; m / s\)

\(s =\frac{1}{2} \times 1.25 \times 8 \times 8\; m / s =40 m\)

now, \(40=-10 t +\frac{1}{2} \times 10 \times t ^2\)

\(5 t^2-10 t-40=0\)

\(t^2-2 t-8=0\)

hence, \(t=4 s\)

\(s=40 \;m\).

displacement \(=40\; m\)

Just after being released the stone has an upward velocity, so it will move upwards first.

distance in an upward direction before stopping \(d\).

\(d =\frac{ v ^2- u ^2}{2 g }=\frac{10^2-0^2}{2 * 10}=5\; m\)

and distance \(= s +2 d =50 m\)

So the distance covered is \(50\; m\) and the displacement is \(40\; m\).