The position of a projectile launched from the origin at $$t = 0$$  is given by $$\vec r = \left( {40\hat i + 50\hat j} \right)m$$    at $$t = 2s.$$  If the projectile was launched at an angle $$\theta $$ from the horizontal, then $$\theta $$ is
(take $$g = 10\,m{s^{ - 2}}$$  )

Solution :
From question,
Horizontal velocity (initial), $${u_x} = \frac{{40}}{2} = 20\,m/s$$
Vertical velocity (initial), $$50 = {u_y}t + \frac{1}{2}g{t^2}$$
$$\eqalign{ & \Rightarrow {u_y} \times 2 + \frac{1}{2}\left( { - 10} \right) \times 4;\,\,{\text{or,}}\,\,{u_y} = \frac{{70}}{2} = 35\,m/s \cr & \therefore \tan \theta = \frac{{{u_y}}}{{{u_x}}} = \frac{{35}}{{20}} = \frac{7}{4} \Rightarrow {\text{Angle}}\,\theta = {\tan ^{ - 1}}\frac{7}{4} \cr} $$