A projectile is projected from  top of tower with velocity 50 m/s at 370 above with horizontal. Velocity of the particle after 1 sec Cos 530=45   g=10m/sec2

given initial velocity U=50 m/s; θ=370; time =t=1 sec;g=10m/s2Ux=U cosθUx=50×45Ux=40m/secVx=Ux+at  ; here a=0 =horizontal accelerationVx=40m/secUy=UsinθUy=50×35=30m/sVy=Uy−gtVy=30−10(1)=20m/secV=Vx2+Vy2=402+202=1600+400=44.72m/sec