A bomb at rest at the summit of a cliff breaks into two equal fragments. One of the fragments attains a horizontal velocity of 20 3ms-1 . The horizontal distance between the two fragments, when their displacement vectors is inclined at 600 relative to each other is (g =10ms-2 )

At time t after the explosion , when displacement vector are at an angle 600 with each other angle with the vertical for each displacement vector  is 300. Then tan300= ut12gt2=2ugt so t=2u3g seperation between them =2ut=2u2u3g=4u23g=4×400×3310                                                                                =4803