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A cracker is thrown into air with a velocity of $10 m/s$ at an angle of $45^{\circ}$ 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 the centre of mass when it is at a height of $1 \;m$ from the ground. $(g=10\;m/s^2)$

\[\begin {array} {1 1} (a)\;4 \sqrt 5\;m/s & \quad (b)\;2 \sqrt 5\; m/s \\ (c)\;5 \sqrt 4 m/s & \quad  (d)\;10 m/s \end {array}\]

1 Answer

Velocity of COM will be the same as that of the projected body not having undergone an explosion.
$V^2= u^2-2gy$
$V^2= 100 -2 \times 10 $
$V= 4 \sqrt 5\;m/s$


answered Nov 22, 2013 by meena.p

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