AB is an inclined plane of inclination 30° with horizontal. Point O is 20m above point A. A particle is projected horizontally and it collides  perpendicular to the plane AB . Speed of the particle, at ‘O’ must be  g=10 m/s2

Resolving initial velocity and  g along  x and  y directionWe get  ux=ucosθ,  uy=usinθ,  θ=30∘gx=−gsinθgy=gcosθFrom diagramApplying equation of motion in  x-directionvx=0 (∵ body hits the plane normally )ux=ucosθ⇒vx=ux+axt⇒0=ucos30°−gsin30°t⇒u3=gt⇒t=3ug……. (1)Applying equation of motion in y direction⇒xy=uyt+12ayt2⇒20cos⁡30∘=usin⁡30∘t+12gcos⁡30∘t2…… (2) ⇒203=ut+12gt23⇒203=u23g+32g⋅3u2g2 (∵g=10m/s)⇒203=u223+332g⇒203=u2532g⇒u2=80Solving the above equation we get  u=45 m/sTherefore, the correct answer is (C).