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Where u and v and initial and final velocities

Looking for some help with another projectiles question:

A particle is projected from the ground with a velocity of 50.96 m/s at an angle Tan-1 5/12 to the horizontal. On its upward path, it just passes over a wall 14.7m high. During its flight it also passes over a second wall, 18.375 m high.
Show that the second wall must not be less than 23.52 m and not more than 70.56 m from the first wall.

A plane is inclined at an angle of 2B (beta) to the vertical. A particle is projected up the plane with initial velocity uCosB, at an angle B (beta) to the inclined plane. The plane of projection is vertical and contains the line of greatest slope.
(I) that the time of flight of the particle is u/g
(II)that the range ofthe particle on the plane is u^2/2g
Ok, so for part (A) ive made this sketch:


Ux = 47.04
Uy = 19.6
Vx = 47.04
Vy = 19.6 - gt
Ax = 0
Ay = -g
Sx = 47.04t
Sy = 19.6t - 1/2gt^2

where u and v and initial and final velocities, and a and s are acceleration and distance, x and y are in the horizontal and vertical directions.

I subbed in Sy = 14.7 and got a quadratic with two solutions - t=1 and t=3, since the time of flight is 8 seconds, either solution is acceptable.
Im stuck as to what to do next. Any hep would be appreciated.

(B) Heres my sketch for this part:

Im totally stuck here. The given velocity is what i would have thought to be the x velocity, suggesting the particle has no y velocity - which cant be possible i think? Any help as to how to break this down would be helpful
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Any help is appreciated

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