\[\begin {array} {1 1} (a)\;30\;m & \quad (b)\;60\;m \\ (c)\;90\;m & \quad (d)\;40\;m \end {array}\]

\[\begin {array} {1 1} (a)\;30\;m & \quad (b)\;60\;m \\ (c)\;90\;m & \quad (d)\;40\;m \end {array}\]

Since centre of mass motion will not be affected by the explosion, it will be at a distance $R$ =range from the point of projection .

=> Total distance of the 2 nd mass from the point of projection is $\large\frac{3R}{2}$

$S= \large\frac{3}{2} \times \frac{u^2}{2g}$

$\quad= \large\frac{3}{2} \times \frac{20 \times 20}{20}$

$\quad= 30 \;m$

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