A stone was vertically projected with a velocity of 20 ms¹ and, 2 seconds later, another stone is projected vertically with a velocity of 20 ms¹. Find the time and distance from the ground when they meet. (g=10ms²)

Let $t$ be the time elapsed since the first stone was projected. The acceleration due to gravity is $g = 10 \text{ m s}^{-2}$ (downwards). We will take the upward direction as positive. The equation for vertical displacement is $h = ut - \frac{1}{2}gt^2$.

For the first stone:
Initial velocity $u_1 = 20 \text{ m s}^{-1}$.
Time in air $= t$.
Height $h_1 = u_1 t - \frac{1}{2}gt^2$

$$h_1 = 20t - \frac{1}{2}(10)t^2$$ $$h_1 = 20t - 5t^2 \quad (*)$$

For the second stone:
Initial velocity $u_2 = 20 \text{ m s}^{-1}$.
The second stone is projected 2 seconds later, so its time in air is $t-2$.
Height $h_2 = u_2 (t-2) - \frac{1}{2}g(t-2)^2$

$$h_2 = 20(t-2) - \frac{1}{2}(10)(t-2)^2$$ $$h_2 = 20(t-2) - 5(t-2)^2 \quad (**)$$

When the stones meet, their heights from the ground are equal, so $h_1 = h_2$.

$$20t - 5t^2 = 20(t-2) - 5(t-2)^2$$

$$20t - 5t^2 = 20t - 40 - 5(t^2 - 4t + 4)$$ $$20t - 5t^2 = 20t - 40 - 5t^2 + 20t - 20$$

Cancel $20t$ and $-5t^2$ from both sides:

$$0 = -40 + 20t - 20$$ $$0 = 20t - 60$$ $$20t = 60$$ $$t = \frac{60}{20}$$ $$t = 3 \text{ s}$$

The stones meet 3 seconds after the first stone was projected.

Substitute $t=3 \text{ s}$ into the equation for $h_1$ (or $h_2$).
Using $h_1 = 20t - 5t^2$:

$$h_1 = 20(3) - 5(3)^2$$ $$h_1 = 60 - 5(9)$$ $$h_1 = 60 - 45$$ $$h_1 = 15 \text{ m}$$

The time when they meet is $\boxed{\text{3 s}}$ and the distance from the ground is $\boxed{\text{15 m}}$.