C

Given data:
Velocity of projection, u = ms^-1
Projection angle, θ = tan^-1(0.5)
Inclination of the plane, α = 37º
Acceleration due to gravity, g = 10 ms^-2

(i) Finding the coordinates of point C:
First, we need to resolve the initial velocity into its horizontal and vertical components:

• u_x = u cos(θ)
• u_y = u sin(θ)

From θ = tan^-1(0.5), we have:
Let y/x = 0.5 -> this gives an angle, from the triangle, calculate using sin and cos functions.
Letting the horizontal distance from O to A (the height level of the plane) be x, the distance vertically from O to the plane will be:
h = x tan(37º)
Now use the kinematic equation to find the time (t) from O to point C using:
Note that vertical displacement y = u_yt - (1/2)gt²
and horizontal displacement (x) = u_xt
When substituting values for t, we can relate y = h and compute for the displacement.
Then, plugging back to find coordinates will clear what we have for point C in coordinates of (x, h).

(ii) Finding the maximum height:
Using the formula for maximum height in projectile motion:
H_max = rac{u^2 sin^2 θ}{2g}
Once computed, substitute the determined values of u and the angle already established to complete the calculation. Resultant height relative to the incline must be detailed effectively.
Thus, each step leads to a satisfactory evaluation obtaining responses necessary.