A cannon ball is fired with a velocity 200 m/sec at an angle of 60° with the horizontal. At the highest point of its flight it explodes into 3 equal fragments, one going vertically upwards with a velocity 100 m/sec, the second one falling vertically downwards with a velocity 100 m/sec. The third fragment will be moving with a velocity

Momentum of ball (mass m) before explosion at the highest point =mvi^=mucos60°i^=m×200×12i^=100 mi^ kgms−1Let the velocity of third part after explosion is V After explosion momentum of system =  P→1+P→2+P→3=m3×100j^−m3×100j^+m3×Vi^By comparing momentum of system before and after the explosion m3×100j^−m3×100j^+m3Vi^=100mi^⇒V=300 m​/​s