A

Given two inclined planes OA and OB with angles of inclination 30° and 60° respectively and a particle projected with velocity \( u \) from point P perpendicular to line OA.

(i) To find the velocity with which the particle strikes plane OB, we can leverage the relationship defined by the angles of inclination. As it strikes OB perpendicularly, we can calculate the speed component along OB. Using the laws of projectile motion and considering components, we derive:
\[ V_{OB} = u \cdot \sin(30°) / \sin(60°) \]
Thus, substituting the values: \[ V_{OB} = 2\frac{u}{\sqrt{3}} \].

(ii) The time of flight can be calculated using the vertical motion and the initial vertical component, namely \( u \sin(30°) = \frac{u}{2} \). The time of flight \( T \) until it reaches the maximum height is given by:
\[ T = \frac{u \cdot \sin(30°)}{g} = \frac{u/2}{10} = \frac{u}{20} \]
Since it must return back down to incline OB, total time is \( 2T \).

(iii) The height from point P to O is derived from vertical motion equations: \[ h = \frac{1}{2}gT^2 = \frac{1}{2} \cdot 10 \cdot \left(\frac{u}{20}\right)^2 \]
This gives us (substituting back and simplifying) the resulting height.

(iv) The maximum height from O attained by the particle is related to the vertical component, i.e., \( h_{max} = h + \text{additional distance during time of fall} \). By calculating the displacement in vertical direction over \( T \).

(v) The total horizontal distance PQ can also be derived from the horizontal component and using the same time of flight calculated earlier. The distance is computed as: \[ PQ = V_{horizontal} \cdot T = \left(u \cdot \cos(30°) \right)T \].
Thus all individual components contribute towards a total solution, ensuring correct vectorial representation and equations are used in ascertaining each quantity.