A particle of mass $$100$$ g is projected at time $$t = 0$$ with a speed $$20$$ m s$$^{-1}$$ at an angle $$45°$$ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time $$t = 2$$ s is found to be $$\sqrt{K}$$ kg m$$^2$$ s$$^{-1}$$. The value of $$K$$ is ______.
(Take $$g = 10$$ m s$$^{-2}$$)

The torque ($$\tau$$) about the origin $$O$$ is produced by the force of gravity ($$mg$$) acting vertically downward.

$$\tau = \text{Force} \times \text{Perpendicular distance} = (mg) \cdot x$$

In projectile motion, the horizontal distance at any time $$t$$ is given by: $$x = (u \cos \theta)t$$

$$\tau = mg(u \cos \theta)t$$

$$L = \int_{0}^{t} \tau \, dt = \int_{0}^{t} mgu \cos \theta \cdot t \, dt$$

$$L = mgu \cos \theta \left[ \frac{t^2}{2} \right]_0^t = \frac{1}{2} mgu \cos \theta \cdot t^2$$

$$L = \frac{1}{2} \times 0.1 \times 10 \times 20 \times \frac{1}{\sqrt{2}} \times (2)^2$$

$$L = 2 \times \frac{20}{\sqrt{2}} = \frac{40}{\sqrt{2}} = 20\sqrt{2} \text{ kg m}^2 \text{ s}^{-1}$$

$$\sqrt{K} = 20\sqrt{2}$$

$$K = (20\sqrt{2})^2$$

$$K = 400 \times 2 = 800$$