A body is projected vertically upwards with velocity $u{\text{ }}m{s^{ - 1}}$. When it returns to the point of projection, its velocity is
$
  {\text{A}}{\text{. u m}}{{\text{s}}^{ - 1}} \\
  {\text{B}}{\text{. 0 m}}{{\text{s}}^{ - 1}} \\
  {\text{C}}{\text{. - u m}}{{\text{s}}^{ - 1}} \\
  {\text{D}}{\text{. }}\sqrt u {\text{ m}}{{\text{s}}^{ - 1}} \\
$

Hint: When the body is thrown upwards under earth’s gravity, it faces deceleration due to the acceleration due to gravity and when it falls down it accelerates and we can describe the motion of the ball using equations of motion.

Complete step by step solution:
We are given a body which is projected in vertically upwards direction with velocity u m/s. We have no air resistance in this case so there is additional deceleration.
It should be noted that the acceleration due to gravity is experienced as deceleration during the upward journey of the object because gravity tends to accelerate in downward direction. So, when the body falls down it accelerates with the same amount as it decelerated due to which initial and final velocity is the same as the body now falls in the same direction as the direction of the acceleration.
We can verify this using the equations of motion:
${{\text{v}}^2} - {u^2} = 2aS$
Now if we consider the complete upward and downward journey of the object then total displacement is zero which means S = 0. Therefore, we have
$
  {{\text{v}}^2} = {u^2} \\
   \Rightarrow {\text{v}} = \pm u \\
$
This equation implies that during the upward journey, v = +u m/s while during the downward journey v = -u m/s
So, the correct answer to the question must be option C.

Note: If there had been air resistance while throwing up the ball, then the upward velocity wouldn’t have been equal to the downward velocity because the air resistance tends to decelerate the body during both upward and downward journey.