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

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

  1. A

    100 ms-1 in the horizontal direction

  2. B

    300 ms-1 in the horizontal direction

  3. C

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

  4. D

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

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    Solution:

    At highest point, pi = pf and perpendicular velocity = 0

    So,       (3m)(100i^)=m(100j^)-m(100j^)+m(v)

                            v=300i^ (of third part)

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