A projectile is launched at time t=0 with velocity=20m/s at an angle 53 with the horizontal . the magnitude of velocity of the projectile at the highest point of the trajectory is

Dear Student,
Given,
initial velocity, u = 20 m/s.
At the highest point the vertical component of the velocity will become 0. The particle has only the horizontal component of velocity which is $u\cos \left({53}^o\right)=20\times \frac{3}{5}=12 m/s$ at an angle $0^o$ wrt the horizontal.
Regards