We're asked to find the time when the velocity vector makes an angle of #30^"o"#, given that it starts at an angle of #60^"o"#.
The initial velocity components are
#v_(0x) = ucos60^"o" = u/2# (this value doesn't change throughout motion)
#v_(0y) = usin60^"o" = (usqrt3)/2#
When the velocity vector is at an angle of #30^"o"# from the horizontal, the horizontal component is still #u/2#, but we need to find the vertical component, using trigonometry.
#tantheta = (v_y)/(v_x)#
#v_y = v_xtan30^"o" = u/2((1)/sqrt3)#
#v_y = color(red)((u)/(2sqrt3)#
Now that we know the #y#-velocity at this time, we can use the equation
#v_y = v_(0y) - g t#
to find the time #t# when this occurs. Plugging in known values, we have
#color(red)((u)/(2sqrt3)) = (usqrt3)/2 - g t#
#t = (color(red)((u)/(2sqrt3)) - (usqrt3)/2)/(-g)#
#t = (-(usqrt3)/3)/(-g)#
#color(blue)(t = (usqrt3)/(3g)#
or
#color(blue)(t = u/(gsqrt3))#
which is approximately
#color(blue)(t ~~ 0.0589u#
