Let the upward direction be positive.

For vertical motion under gravity, position after time $t$ is $y = ut - \frac{1}{2}gt^2$

Position of the first ball

The first ball is thrown at $t=0$ with speed $50\,\text{m/s}$.

So at time $t$, its position is $y_1 = 50t - \frac{1}{2}(10)t^2 = 50t - 5t^2$

Position of the second ball

The second ball is thrown at $t=2\,\text{s}$ with the same speed $50\,\text{m/s}$.

For $t \ge 2$, its time of flight is $(t-2)$, so its position is $y_2 = 50(t-2) - 5(t-2)^2$

Condition for meeting

They meet when their positions are equal: $y_1 = y_2$

So, $50t - 5t^2 = 50(t-2) - 5(t-2)^2$

Expand and simplify

First expand the right side: $50(t-2) - 5(t-2)^2 = 50t - 100 - 5(t^2 - 4t + 4)$ $= 50t - 100 - 5t^2 + 20t - 20$ $= 70t - 120 - 5t^2$

Now equate: $50t - 5t^2 = 70t - 120 - 5t^2$

The $-5t^2$ terms cancel: $50t = 70t - 120$ $20t = 120$ $t = 6\,\text{s}$

Final answer

The second ball meets the first ball at $\boxed{6\ \text{s}}$

Comparison with stored answer

Stored correct answer = $6$

Our derived answer also is $6$, so they agree.