Two particles are simultaneously thrown from the roofs of two high buildings as shown in figure. Their velocities are  VA= 2 m/s  and VB=14 m/s respectively. Then

Assuming A to at rest  a¯BA=a¯B−a¯A=0 as a¯A=a¯B−g  (downwards)Thus, the relative motion between them is uniform. Relative velocity of B with respect to A in vertical direction, and horizontal directionRelative velocity of B with respect to A in horizontal direction.uBAV= uBsin 45-uAsin 45 = 1212=62m/sRelative velocity of B with respect to A in vertical direction.uBAH= uBcos 450 - (-uAcos 450)=14+212=82m/s tanθ=UABVUABH=6282=34 θ=37°Horizontal distance between A and B after time t isx=22−(uBAH)t=(22−82t)mand vertical distance between A and B time t isy=9−uBAVt=9−62tmTherefore, distance between them after time t is  s=x2+y2For S to be minimum  ddt(s2)=0∴(22−82t)(−82)+2(9−62t)(−62)=0Therefore   t=23102Stan37=x+922 34=x+922 x=7.5m sin53=x'x 45=x'7.5 x'=6mParagraph- IA particle is projected horizontally with a speed V = 5 m/s from the top of a plane inclined at an angle θ=370  to the horizontal. (g = 10 m/s2)