Question

Two balls $A$ and $B$ of equal masses are projected upward simultaneously, one from the ground with speed $50m{s}^{-1}$ and other from height $40m$ above the first ball high tower with initial speed $30m{s}^{-1}$. Find the maximum height attained by their centre of mass.

Answer 1

The initial position of centre of mass is
$y_{ic}=\frac{m\times 0+m\times 40}{2m}=20m$
Initial velocity of centre of mass
$u_{cm}=\frac{m\times 50+m\times 30}{2m}=40m{s}^{-1}$
Acceleration of centre of mass, $a_{cm}=-g$
Using kinematic equation: ${\nu }_{cm}^2=u_{cm}^2+2a_{cm}H$
here $H$ is the maximum height reach by centre of mass of two balls from initial level.
$\therefore 0^2={40}^2-2\times 10\times H$
$⟹H=\frac{1600}{20}=80m$
Hence, maximum height reach by centre of mass from ground level will be
$h_{max}=(h_{cm}{)}_{initial}+H=20+80=100m$