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

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Infinity Learn · Accepted Answer

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

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