Two balls of equal masses are projected upwards simultaneously, one from the ground with initial velocity $50{\text{ms}}^{-1}$, and the other from a $40\text{m}$ tower with initial velocity of $30{\text{ms}}^{-1}.$The maximum height attained by their COM will be:
($g=10{\text{ms}}^{-2}$)

The correct option is C $100\text{m}$

Initially, assuming the bottom of the tower to be origin and upward direction to be $+ve$ - $y$ axis

$y_{com}=\frac{m\left(0\right)+m\left(40\right)}{2m}=20\text{m}$

Initially, velocity of centre of mass of the system is given by,


$u_{com}=\frac{m\left(50\right)+m\left(30\right)}{2m}=40{\text{ms}}^{-1}$

$a_{com}=-g=-10{\text{ms}}^{-2}$

$v_{com}^2-u_{com}^2=2a_{com}H$

$0=1600-2\times 10\times H$

$\Rightarrow H=80\text{m}$

$\Rightarrow y_f-y_i=80\text{m}\Rightarrow y_f=80+20=100\text{m}$

Hence, option $\left(C\right)$ is the correct answer.