C

Step 1: When a particle is projected with an initial velocity \( u \) at an angle \( \theta \), its horizontal and vertical components of velocity can be expressed as:
\( u_x = u \cos \theta \) (horizontal component)
\( u_y = u \sin \theta \) (vertical component)

Step 2: The velocity \( v \) at any instant can be described as a vector sum of these components due to gravity acting vertically downward.

Step 3: As per the question, the velocity \( v \) is at right angles to the initial velocity \( u \); this indicates that the two vectors are perpendicular.

Step 4: For two vectors to be perpendicular, their dot product must be zero. Thus, we can find the relationship between the components.

Step 5: Since \( v \) is at right angles to \( u \), using trigonometric identities, we can express \( v \) as \( v = \sqrt{u_x^2 + u_y^2} \). Given that \( u_y = u \sin \theta \) and \( u_x = u \cos \theta \), we can relate them to form the required equation due to being perpendicular.

Step 6: The magnitude of the velocity when it is perpendicular will be \( u_y = u \sin \theta \) on vertical and converting horizontal component, we find:
\( u_y = u \tan \theta \cdot \cos \theta \) leading us to eventual relations of angles and angles leading to \( u \cot \theta \) when the particle moves to its maximum height.

Therefore, the value of the velocity \( v \) is given by \( \mathbf{u \cot \theta} \). Thus, the correct option is C.