A particle of mass 2 kg is moving such that at time t, its position, in meter, is given by $$\vec{r}(t) = 5\hat{i} - 2t^2\hat{j}$$. The angular momentum of the particle at $$t = 2s$$ about the origin in kgm$$^{-2}$$ s$$^{-1}$$ is :

The angular momentum of a particle about a point is given by the cross product of its position vector and its linear momentum, that is, $$\vec{L} = \vec{r} \times \vec{p}$$. Here, $$\vec{p} = m \vec{v}$$, where $$m$$ is the mass and $$\vec{v}$$ is the velocity vector.

First, we are given the position vector as a function of time: $$\vec{r}(t) = 5\hat{i} - 2t^2\hat{j}$$. The mass $$m = 2$$ kg.

To find the velocity $$\vec{v}(t)$$, we differentiate the position vector with respect to time:

$$\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(5)\hat{i} + \frac{d}{dt}(-2t^2)\hat{j}$$

The derivative of the constant 5 is 0, and the derivative of $$-2t^2$$ is $$-4t$$. So,

$$\vec{v}(t) = 0 \cdot \hat{i} - 4t \hat{j} = -4t \hat{j} \text{ m/s}$$

Now, the linear momentum $$\vec{p} = m \vec{v} = 2 \times (-4t \hat{j}) = -8t \hat{j} \text{ kg m/s}$$.

Next, we compute the angular momentum $$\vec{L} = \vec{r} \times \vec{p}$$:

$$\vec{L} = (5\hat{i} - 2t^2\hat{j}) \times (-8t \hat{j})$$

Expanding the cross product:

$$\vec{L} = 5\hat{i} \times (-8t \hat{j}) + (-2t^2\hat{j}) \times (-8t \hat{j})$$

Let's evaluate each term separately. For the first term:

$$5\hat{i} \times (-8t \hat{j}) = 5 \times (-8t) \times (\hat{i} \times \hat{j}) = -40t \times \hat{k} = -40t \hat{k}$$

because $$\hat{i} \times \hat{j} = \hat{k}$$. For the second term:

$$(-2t^2\hat{j}) \times (-8t \hat{j}) = (-2t^2) \times (-8t) \times (\hat{j} \times \hat{j}) = 16t^3 \times \vec{0} = 0$$

since the cross product of any vector with itself is zero. Therefore,

$$\vec{L} = -40t \hat{k} + 0 = -40t \hat{k} \text{ kg m}^2\text{s}^{-1}$$

Now, we need the angular momentum at $$t = 2$$ s. Substituting $$t = 2$$:

$$\vec{L} = -40 \times 2 \hat{k} = -80 \hat{k} \text{ kg m}^2\text{s}^{-1}$$

Comparing with the given options, $$-80\hat{k}$$ corresponds to option A.

Hence, the correct answer is Option A.