If x and y coordinates of a projectile as a function of time (t) are given as $$24t$$ and $$43.6t - 4.9t^2$$, respectively, then the angle (in degrees) made by the projectile with horizontal when $$t = 2$$ s is ______.

The position of the projectile is given as a function of time:

$$x = 24t, \quad y = 43.6t - 4.9t^{2}$$

The instantaneous velocity components are obtained by differentiating the position coordinates with respect to time.

Horizontal component: $$v_x = \frac{dx}{dt} = \frac{d(24t)}{dt} = 24 \text{ m\,s}^{-1}$$

Vertical component: $$v_y = \frac{dy}{dt} = \frac{d\bigl(43.6t - 4.9t^{2}\bigr)}{dt} = 43.6 - 9.8t \text{ m\,s}^{-1}$$

At $$t = 2$$ s, substitute in the expressions above:

$$v_x = 24 \text{ m\,s}^{-1}$$ (constant for all $$t$$)
$$v_y = 43.6 - 9.8(2) = 43.6 - 19.6 = 24 \text{ m\,s}^{-1}$$

The angle $$\theta$$ that the velocity vector makes with the horizontal is given by

$$\tan\theta = \frac{v_y}{v_x} = \frac{24}{24} = 1$$

Therefore, $$\theta = 45^{\circ}$$.

Hence, the correct choice is:
Option B which is: $$45°$$