A particle of mass m is flying horizontally at  velocity u. It strikes a smooth inclined surface and  its velocity becomes vertical.

The inclination of the incline to the horizontal is $\theta ={60}^0.$Then

 

 

(a) Since the incline surface is smooth, the velocity component of the particle parallel to the incline will not  change.
 

$$\begin{gathered} \therefore \mathrm{Vsin}\mathrm{\theta }=\mathrm{ucos}\mathrm{\theta }\dots \dots \dots \left(1\right) \\ \text{ For }\mathrm{\theta }={60}^{\circ },\mathrm{V}=\frac{\mathrm{u}}{\sqrt{3}} \\ \text{ Loss in }\mathrm{KE}=\frac{1}{2}{\mathrm{mu}}^2-\frac{1}{2}\mathrm{m}{\left(\frac{\mathrm{u}}{\sqrt{3}}\right)}^2=\frac{{\mathrm{mu}}^2}{3} \\ \text{ (b) From (1) }\tan \mathrm{\theta }=\frac{\mathrm{u}}{\mathrm{V}} \\ \text{ Since, }\mathrm{u}>\mathrm{V}\therefore \tan \mathrm{\theta }>1\therefore \mathrm{\theta }>{45}^{\circ } \end{gathered}$$
Therefore, particle cannot go vertically up if $\mathrm{\theta }<{45}^{\circ }.$