TS EAMCET 9-Sep-2020 Shift 1 Solved Paper

As second ball is dropped 0.2 s after first ball, we have following situation at t=0.4s .Now, y1= distance travelled by first ball in 0.4s
=‌

1

2

gt2=‌

1

2

×10×(0.4)2=0.8m
and y2= distance travelled by second ball in 0.2s
=‌

1

2

gt2=‌

1

2

×10×(0.2)2=0.2m
Now, position of centre of mass of ball w.r.t. release point (taken origin) is
yCM=‌

m1y1+m2y2

m1+m2

Hence, distance between centre of mass and release point is 0.4m.