a particle is projected at an initial velocity u with a angle of 60 with the vertical. The time after which the velocity vector becomes perpendicular to the initial velocity vector?

final angle of velocity vector is 30 with horizontal

show their horizontal components of velocity same and by vertical motion calc time

When a projectile makes 60⁰ it makes an angle 30⁰ with the horizontal.the velocity vector of the partical initially is v₀=ucos30i+usin30j, after a time t the velocity vector is v=ucos30i+(usin30-gt)j the two vectors are perpendicular so their dot product is zero. we get that ucos30xucos30+usin30xusin30=gtsin30u=>u²=gtsin30=>t=u/gsin30

(ucos30°i + (usin30° - gt)j).(ucos30°i + usin30°j) = 0

u² = ugtsin30°

t = 2u/g.

this is the time after which the velocity of the particle becomes perpendicular to the initial velocity vector.