A cannon ball is fired with a velocity 200 ms-1 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 ms-1, the second one falling vertically downwards with a velocity 100 ms-1. The third fragment will be moving with a velocity

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100 ms-1 in the horizontal direction

300 ms-1 in the horizontal direction

300 ms-1 in a direction making an angle of 60° with the horizontal

200 ms-1 in a direction making an angle of 60° with the horizontal

answer is B.

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Detailed Solution

At highest point, pi = pf and perpendicular velocity = 0So, (3m)(100i^)=m(100j^)-m(100j^)+m(v)∴ v=300i^ (of third part)

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