Two balls A and B of equal masses are projected upwards simultaneously, one from the ground with speed $50 {\mathrm{ms}}^{-1}$ and other from a tower of height $40 \mathrm{m}$ above the first ball with initial speed $30 {\mathrm{ms}}^{-1}\text{.}$ Calculate the maximum height attained by their centre of mass in metre. $\left[\mathrm{Take} \mathrm{g}=10{\mathrm{ms}}^{-2}\right]$

Let us consider ball A and B as a system, then the initial height of centre of mass from ground is

${\left({\mathrm{y}}_{\mathrm{cm}}\right)}_{\mathrm{i}}=\frac{\mathrm{m}\left(0\right)+\mathrm{m}\left(40\right)}{2\mathrm{m}}=20\mathrm{m}$

Initial velocity of centre of mass is

${\mathrm{u}}_{\mathrm{cm}}=\frac{\mathrm{m}\left(50\right)+\mathrm{m}\left(30\right)}{2\mathrm{m}}=40{\mathrm{ms}}^{-1}$

Acceleration of centre of mass is

${\mathrm{a}}_{\mathrm{cm}}=-\mathrm{g}=-10{\mathrm{ms}}^{-2}$

$\text{ Since }{\mathrm{v}}_{\mathrm{cm}}^2-{\mathrm{u}}_{\mathrm{cm}}^2=2{\mathrm{a}}_{\mathrm{cm}}{\mathrm{\Delta y}}_{\mathrm{cm}}$

Where ${\mathrm{\Delta y}}_{\mathrm{cm}}$ is displacement of centre of mass along y direction (taken vertically upwards)

$\Rightarrow 0^2-{40}^2=-2\times 10\times \mathrm{\Delta }{y}_{\mathrm{cm}}$

$\Rightarrow {\mathrm{\Delta y}}_{\mathrm{cm}}=\frac{1600}{20}=80\mathrm{m}$

$\Rightarrow {\left({\mathrm{y}}_{\mathrm{cm}}\right)}_{\mathrm{f}}-{\left({\mathrm{y}}_{\mathrm{cm}}\right)}_{\mathrm{i}}=80$

$\Rightarrow {\left({\mathrm{y}}_{\mathrm{cm}}\right)}_{\mathrm{f}}=20+80=100\mathrm{m}$

Hence, maximum height reach by centre of mass from ground level will be

${\mathrm{h}}_{max}=100\mathrm{m}$