Oblique projectile From Inclined Surface

v) Velocity of the Projectile at any Instant 't':

The horizontal component of the projectile remains constant all the time. (because acceleration due to gravity has no component along the horizontal.)
.. Horizontal component of velocity after any time t is v_x = u_x = ucos\(\theta\)
Vertical component of velocity after any time t is v_y=u_y-gt=usin\(\theta\) -gt.
                      \( \overrightarrow V = V_x \hat{i} + V_y \hat{j} \)
        \( \overrightarrow V = (u\cos \theta )\hat{i} + \)(usin\(\theta\)-gt)\(\hat j\)
Then the magnitude of resultant velocity after time t is
             \( V = \sqrt {V_x ^2 + V_y ^2 } = \sqrt {(u\cos \theta )^2 + \left( {u\sin \theta - gt} \right)^2 } \)
The velocity vector v makes an angle a with the horizontal given by
       \(\theta\) = \( \tan ^{ - 1} \left( {\frac{{V_y }} {{V_x }}} \right) \)  at this instant
          
     
At any vertical displacement 'h', velocity is 
         \( V = u\cos \theta \hat i + \sqrt {u^2 \sin ^2 \theta - 2gh} \hat j \)