Again stones?

Let the dropping stone cover a distance of $x$ m before the collision i.e. the collision happen at $(80-x)$ m above ground level.

For the dropping stone :-
$\begin{aligned} S & = ut + \dfrac{1}{2}{at}^2\\ \\ x & = 0×t + \dfrac{1}{2} × 10 × t^2\\ \\ x & = 5t^2 \longrightarrow \boxed{1} \end{aligned}$

For thrown-up stone :-
$\begin{aligned} S & = ut + \dfrac{1}{2}{at}^2\\ \\ 80-x & = 20×t - \dfrac{1}{2}×10×t^2 \left(\text{because acceleration here is -g}\right)\\ \\ 80-x & = 20t - 5t^2\\ \\ \text{Adding} \ \boxed{1} \ \text{and} \ \boxed{2}:-\\ \\ 80 & = 20t\\ \\ t & = \color{#3D99F6}{\boxed{4.00}} \end{aligned}$

Alternatively:-
Rotate the figure ${90}^°$ left:-

We observe that the relative velocity is $20 - 0 = 20m/s$ , relative acceleration is $0m/s^2$ and distance is $80m$ .
So, time = $\dfrac{\text{distance}}{\text{relative velocity}} = \dfrac{80}{20} = \color{#3D99F6}{\boxed{4.00}}$

Let us first consider the stone that I had dropped. Let $x$ be its distance above the ground, $v$ be its velocity, $t$ be the time elapsed. We have $v=-gt\\\frac{dx}{dt}=-gt\\\int dx=\int20-gt\space dt\\x=-\frac{gt^2}{2}+C=-\frac{gt^2}{2}+80$

Let us now consider the stone that was thrown up. Let $x$ be its distance above the ground, $v$ be its velocity, $t$ be the time elapsed. We have $v=20-gt\\\frac{dx}{dt}=20-gt\\\int dx=\int20-gt\space dt\\x=20t-\frac{gt^2}{2}$

Equating the two, we have $20t-\frac{gt^2}{2}=-\frac{gt^2}{2}+80\\\therefore t=4.00$

We can model the vertical positions of the stones as functions of time.

My stone is $y_1(t)=80-5t^2$

Abhay's stone is $y_2(t)=20t-5t^2$

We can set them equal to each other and solve for t.

$80-5t^2=20t-5t^2$

$t=4 s$