A ship A is moving Westwards with a speed of 10 km h^$$-$$1 and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h^$$-$$1. The time after which the distance between them becomes shortest, is

A projectile is fired from the surface of the earth with a velocity of 5 m s^$$-$$1 and angle $$\theta $$ with the horizontal. Another projectile fired from another planet with a velocity of 3 m s^$$-$$1 at the same angle follows a trajectory which is identical with the trajectory of the projectile fired from the earth. The value of the acceleration due to gravity on the planet is (in m s^$$-$$2) is
(Given g = 9.8 m s^$$-$$2)

A particle is moving such that is position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m)
at time t = 2 s and (13 m, 14 m) at time t = 5 s.

Average velocity vector $$\left( {{{\overrightarrow v }_{av}}} \right)$$ from t = 0 to t = 5 s is

Vectors $$\overrightarrow A ,\overrightarrow B $$ and $$\overrightarrow C $$ are such that $$\overrightarrow A .\overrightarrow B = 0$$ and $$\overrightarrow A .\overrightarrow C = 0$$. Then the vector parallel to $$\overrightarrow A $$ is