A particle shows distance - time curve as given in this figure. The maximum instaneous velocity of the particle is around the point

$$\overrightarrow A $$ and $$\overrightarrow B $$ are two vectors and $$\theta $$ is the angle between them, if $$\left| {\overrightarrow A \times \overrightarrow B } \right| = \sqrt 3 \left( {\overrightarrow A .\overrightarrow B } \right),$$ the value of $$\theta $$ is

A particle starting from the origin (0, 0) moves in a straight line in the (x, y) planes. Its coordinates at a later time are $$\left( {\sqrt 3 ,3} \right)$$. The path of the particle makes with the x-axes an angle of

For angles of projection of a projectile at angle (45^o $$-$$ $$\theta $$) and (45^o + $$\theta $$), the horizontal range described by the projectile are in the ratio of