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A firecracker is thrown with velocity of $30 \mathrm{~ms}^{-1}$ in a direction which makes an angle of $75^{\circ}$ with the vertical axis. At some point on its trajectory, the firecracker splits into two identical pieces in such a way that one piece fall 27 $\mathrm{m}$ far from the shooting point. Assuming that all trajectories are contained in the same plane, how far will the other piece fall from the shooting point ? (Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ and neglect air resistance)-
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The correct answer is:
$63 \mathrm{~m}$ or $117 \mathrm{~m}$
Undér influéncé of constānt forcee céntre of māss follows its original path
$\begin{array}{l}
\mathrm{R}=\frac{30 \times 30 \times \frac{1}{2}}{10}=45 \mathrm{~m} \\
45=\frac{\pm \mathrm{m} \times 27+\mathrm{mx}}{\mathrm{m}+\mathrm{m}} \\
\mathrm{x}=63 \mathrm{~m}, 117 \mathrm{~m}
\end{array}$
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