Oblique Projectile From Inclined Surface

THE TRAJECTORY OF PROJECTILE IS A PARABOLA
(In absence of air resistance)

In absence of air resistance)
Let a body be projected at 'O' with an initial velocity u that makes an angle
This velocity can be written as with the X-axis.
                \( \overrightarrow u = u_x \hat{i} + u_y \hat{j} = (u\cos \theta )i + (u\sin \theta )j \) 

Due to the fact that two dimensional motion can be treated as two independent rectilinear motions, the projectile motion can be broken up into two separate straight line motions.
i) horizontal motion with zero acceleration. [i.e... constant velocity as there is no force in horizontal direction]
ii) vertical motion with constant downward acceleration = g
(it is moving under gravity)
                           
       
Let the body reach a point 'P (x, y)' in its trajectory after time 't' i.e, horizontal displacement and vertical displacements of the body in time 't' are x and y respectively.
1) Let us first consider the horizontal motion. As the horizontal motion has no acceleration the horizontal component of projectile's velocity u, remains constant through out the motion, the displacement of the projectile after any time 't' from the initial position (the origin in our case) is given by
             \( x = u_t + \frac{1} {2}a_x t^2 \)
Since horizontal acceleation is zero,
        \( x = u_x t = (u\cos \theta )t \).............(1)