A projectile is thrown upward at an angle $$60 ^{o}$$ with the horizontal. The speed of the projectile is 20 m/s when its direction of motion is $$45 ^{o}$$ with the horizontal. The initial speed of the projectile is ______ m/s.

We need to find the initial speed of a projectile thrown at $$60°$$ to the horizontal. The angle of projection is $$\theta = 60°$$, and at some point the speed is $$20$$ m/s when the direction of motion is $$45°$$ with the horizontal.

The horizontal component of velocity remains constant throughout the motion, given by $$v_x = u\cos 60° = \frac{u}{2}$$. At the point where the direction is $$45°$$, the horizontal component is also $$v_x = 20\cos 45° = \frac{20}{\sqrt{2}} = 10\sqrt{2}$$.

Equating the two expressions for the horizontal component gives $$\frac{u}{2} = 10\sqrt{2}$$, from which it follows that $$u = 20\sqrt{2}$$.

Therefore, the initial speed is $$20\sqrt{2}$$ m/s, which matches Option D.