A stone is dropped from the top of a tower of height 60 m. At the same time, another stone is thrown up from the foot of the tower with a velocity 20 m/s. Calculate the displacement of the centre of mass of the two stones at time of collision. (g=10 m/`s^(2))`

Explanation

Initially the centre of mass of the two stones is at a height of 30 m from the ground. Acceleration of the centre of mass
`= (mxxg+mxxg)/(m+m)`=g (downward)
Let the stones colide after time t.
At that time distance travelled by the first stone `= (1)/(2) "gt"^(2)`
and distance travelled by the second stone = `ut -(1)/(2)"gt"^(2)`
Now, `(1)/(2)"gt"^(2)+ut-(1)/(2)"gt"^(2)=60`
`:." "t=(60)/(u)=(60)/(20)=3 s`
Initital speed of the centre of mass,
`u_("cm") = (mxx0+mxx20)/(m+m)` = 10 m/s (upward)
`:.` Displacement of the centre of mass
`= u_("cm")xxt-(1)/(2)"gt"^(2)=10xx3-(1)/(2)xx10xx9=-15`m
Therefore, the centre of mass will move downward by 15 m.