Given initial velocity u = (20sqrt(3)i + 20j) ms⁻¹.

The horizontal velocity remains constant: $v_x = u_x = 20\sqrt{3}$ ms$^{-1}$.
The vertical velocity changes due to gravity: $v_y = u_y - gt$.
At $t=1$ s:

$$v_x = 20\sqrt{3}\, \text{ms}^{-1}$$ $$v_y = 20 - (10)(1) = 10\, \text{ms}^{-1}$$

So, the velocity vector at $t=1$ s is $\mathbf{v} = (20\sqrt{3}\mathbf{i} + 10\mathbf{j})$ ms$^{-1}$.

$$|\mathbf{v}| = \sqrt{v_x^2 + v_y^2}$$ $$|\mathbf{v}| = \sqrt{(20\sqrt{3})^2 + (10)^2}$$ $$|\mathbf{v}| = \sqrt{(400 \times 3) + 100}$$ $$|\mathbf{v}| = \sqrt{1200 + 100}$$ $$|\mathbf{v}| = \sqrt{1300} = \sqrt{100 \times 13} = 10\sqrt{13}\, \text{ms}^{-1}$$

The direction is given by the angle $\alpha$ with the horizontal.

$$\tan \alpha = \frac{v_y}{v_x} = \frac{10}{20\sqrt{3}} = \frac{1}{2\sqrt{3}}$$ $$\alpha = \arctan\left(\frac{1}{2\sqrt{3}}\right) \approx 16.1^\circ$$

The speed of the particle after 1 second is $\boxed{10\sqrt{13}\, \text{ms}^{-1}}$ and its direction is $\boxed{\arctan\left(\frac{1}{2\sqrt{3}}\right) \text{ above the horizontal}}$.

The vertical displacement is given by $y = u_y t - \frac{1}{2}gt^2$.
Set $y=0$:

$$0 = u_y T - \frac{1}{2}gT^2$$ $$0 = T\left(u_y - \frac{1}{2}gT\right)$$

This gives $T=0$ (initial projection) or $u_y - \frac{1}{2}gT = 0$.

$$\frac{1}{2}gT = u_y$$ $$T = \frac{2u_y}{g}$$

$$T = \frac{2 \times 20}{10} = \frac{40}{10} = 4\, \text{s}$$

The time of flight of the particle is $\boxed{4\, \text{s}}$.

$$R = (20\sqrt{3})(4) = 80\sqrt{3}\, \text{m}$$

The range of the particle is $\boxed{80\sqrt{3}\, \text{m}}$.

We can use the kinematic equation $v_y^2 = u_y^2 - 2gH$, where $H$ is the maximum height.

$$0^2 = (20)^2 - 2(10)H$$ $$0 = 400 - 20H$$

$$20H = 400$$ $$H = \frac{400}{20} = 20\, \text{m}$$

The maximum height attained by the particle is $\boxed{20\, \text{m}}$.

The horizontal displacement is $x = u_x t$.

$$t = \frac{x}{u_x} = \frac{x}{20\sqrt{3}}$$

$$y = u_y t - \frac{1}{2}gt^2$$ $$y = (20)\left(\frac{x}{20\sqrt{3}}\right) - \frac{1}{2}(10)\left(\frac{x}{20\sqrt{3}}\right)^2$$

$$y = \frac{x}{\sqrt{3}} - 5\left(\frac{x^2}{400 \times 3}\right)$$ $$y = \frac{x}{\sqrt{3}} - 5\left(\frac{x^2}{1200}\right)$$ $$y = \frac{x}{\sqrt{3}} - \frac{x^2}{240}$$

The Cartesian equation of the path of the particle is