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