The position of a projectile launched from the origin at t = 0 is given by $$\vec{r} = (40\hat{i} + 50\hat{j})$$ m at t = 2s. If the projectile was launched at an angle $$\theta$$ from the horizontal, then $$\theta$$ is (take g = 10 ms$$^{-2}$$).

The position vector at time $$ t = 2 $$ seconds is given as $$ \vec{r} = 40\hat{i} + 50\hat{j} $$ meters. This means the horizontal displacement $$ x = 40 $$ m and the vertical displacement $$ y = 50 $$ m.

For a projectile launched from the origin with initial velocity $$ u $$ at an angle $$ \theta $$ to the horizontal, the equations of motion are:

Horizontal motion (no acceleration): $$ x = u_x t $$, where $$ u_x = u \cos \theta $$ is the horizontal component of initial velocity.

Vertical motion (constant acceleration due to gravity $$ g = 10 $$ m/s² downward): $$ y = u_y t - \frac{1}{2} g t^2 $$, where $$ u_y = u \sin \theta $$ is the vertical component of initial velocity.

At $$ t = 2 $$ s, $$ x = 40 $$ m. Substituting into the horizontal equation:

$$ 40 = u_x \cdot 2 $$

Solving for $$ u_x $$:

$$ u_x = \frac{40}{2} = 20 \text{ m/s} $$

At $$ t = 2 $$ s, $$ y = 50 $$ m. Substituting into the vertical equation:

$$ 50 = u_y \cdot 2 - \frac{1}{2} \cdot 10 \cdot (2)^2 $$

First, compute the term $$ \frac{1}{2} \cdot 10 \cdot 4 $$:

$$ \frac{1}{2} \cdot 10 = 5, \quad 5 \cdot 4 = 20 $$

So the equation becomes:

$$ 50 = 2u_y - 20 $$

Add 20 to both sides:

$$ 50 + 20 = 2u_y $$

$$ 70 = 2u_y $$

Solving for $$ u_y $$:

$$ u_y = \frac{70}{2} = 35 \text{ m/s} $$

The angle $$ \theta $$ is related to the components by:

$$ \tan \theta = \frac{u_y}{u_x} $$

Substituting the values:

$$ \tan \theta = \frac{35}{20} = \frac{7}{4} $$

Therefore, $$ \theta = \tan^{-1} \left( \frac{7}{4} \right) $$.

Comparing with the options:

A. $$ \tan^{-1} \frac{3}{2} $$

B. $$ \tan^{-1} \frac{2}{3} $$

C. $$ \tan^{-1} \frac{7}{4} $$

D. $$ \tan^{-1} \frac{4}{5} $$

Option C matches our result.

Hence, the correct answer is Option C.