Question

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

• A. 80 m
• B. 100 m
• C. 20 m
• D. 60 m

Answer 1

Answer: (B)

100 m

The initial position of centre of mass is
$y_{cm}=\frac{m\times 0+m\times 40}{2m}=20m$
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 kinematics equation: $v_{cm}^2=u_{cm}^2+2a_{cm}H$
Here H is the maximum height reach by centre of mass of two balls from initial level.
$\therefore$ $0^2={40}^2-2\times 10\times H$
$\Rightarrow H=\frac{1600}{20}=80m$
Hence, maximum height reach by centre of mass from ground level will be
$h_{max}=(h_{cm}{)}_{initial}+H=20+80=100m$