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Question

# A cracker is thrown into air with a velocity of 10 m/s at an angle of 45o with the vertical. When it is at a height of 0.5 m from the ground, it explodes into a number of pieces which follow different parabolic paths. What is the velocity of centre of mass, when it is at a height of 1 m from the ground? (g=10m/s2)

A
45m/s
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B
25m/s
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C
44m/s
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D
10 m/s
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Solution

## The correct option is A 4√5m/sSince cracker explodes in mid air it does not encounter any external force so velocity of COM will remain unchanged due to explosion.Horizontal component =10sin45o=10√2=5√2m/sFrom third equation of motion, at a height of 1 m, vertical component of velocity of center of mass =√(5√2)2−2×10×1=√50−20=√30m/sHence, net velocity =√(5√2)2+(√30)2=√80=4√5m/s

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