A projectile is projected at 30$$^\circ$$ from horizontal with initial velocity 40 m s$$^{-1}$$. The velocity of the projectile at $$t = 2$$ s from the start will be:

Given a projection angle of 30° and an initial velocity of 40 m/s, after 2 s the horizontal component of velocity remains constant, so $$v_x = u\cos 30° = 40 \times \frac{\sqrt{3}}{2} = 20\sqrt{3} \text{ m/s}$$.

Next, the vertical component of velocity at t = 2 s is $$v_y = u\sin 30° - gt = 40 \times \frac{1}{2} - 10 \times 2 = 20 - 20 = 0 \text{ m/s}$$.

Substituting these into the expression for resultant velocity gives $$v = \sqrt{v_x^2 + v_y^2} = \sqrt{(20\sqrt{3})^2 + 0^2} = 20\sqrt{3} \text{ m/s}$$. At t = 2 s, the projectile is at its maximum height (vertical velocity = 0), so the resultant velocity equals the horizontal component.