A particle is projected from a point $O$ with velocity $u$ at an angle of $60^{\circ }$ with the horizontal. When it is moving in a direction at right angles to its direction at $O$, its velocity then is given by:

The horizontal component of velocity remains constant throughout the flight:$u_x=u\cos 60^{\circ }=u/2$.Let the velocity at the later point be $v$. The angle it makes with the horizontal is $\varphi$.Since the direction is at right angles to $60^{\circ }$, it must be $90^{\circ }-60^{\circ }=30^{\circ }$ below the horizontal.Horizontal component at this point: $v\cos 30^{\circ }$.$v\cos 30^{\circ }=u\cos 60^{\circ }$$v(\sqrt{3}/2)=u(1/2)$$v\sqrt{3}=u⟹v=u/\sqrt{3}$.