A particle is projected with an initial velocity of 40 m/s in a direction making an angle 60 degree above horizontal. find the time after which the velocity of the particle is in a direction making an angle 45 degree above horizontal.

Dear Student,
the horizontal velocity in a projectile motion always remain constant
40cos60=ucos45
=>u=20 m/s is the velocity when the angle with horizontal is 45⁰.

Hence only the vertical component of the velocity changes

Veritcal initial velocity=u1=40sin60
vertical velocity when the hoz angle is 45⁰ =v1= 20 sin45

By equation of motion,
v=u+at
=>v1=u1-gt
=>20 -20=gt
=>t=1.5s