The angle which the velocity vector of a projectile thrown with a velocity $$v$$ at an angle $$\theta $$ to the horizontal will make with the horizontal after time $$t$$ of its being thrown up is:

Solution :
Horizontally after time $$t$$
$$\eqalign{ & v\cos \theta = v\cos \beta \,......\left( {\text{i}} \right)\,\,\left[ {\beta = {\text{angle with horizontal after time}}\,t} \right] \cr & {\text{Vertically}}, \cr & v\sin \theta - gt = v\sin \beta \,.......\left( {{\text{ii}}} \right) \cr} $$
Dividing equation $$\frac{{\left( {{\text{ii}}} \right)}}{{\left( {\text{i}} \right)}}$$ we get
$$\tan \beta = \frac{{v\sin \theta - gt}}{{v\cos \theta }} = \beta = {\tan ^{ - 1}}\left( {\frac{{v\sin \theta - gt}}{{v\cos \theta }}} \right)$$