Two particles A and B of mass 1 kg and 2 kg respectively are projected in the direction shown in figure with speeds `u_(A) = 200 ms^(-1)` and `u_(B) = 50 ms^(-1)`. Initially they were 90 m apart. Find the maximum height attained by the center of mass of the particles. Assume acceleration due to gravity to be constant. (Take, `g=10 ms^(-2)`.

Explanation

Using `m_(A)r_(A)=m_(B)r_(B)`
or `(1)(r_(A))=(2)(r_(B))`

or `r_(A)=2r_(B)` ….(i)
and `r_(A)+r_(B)=90` m …(ii)
Solving these two equations, we get
`r_(A)=60 `m and `r_(B)=30` m
i.e., COM is at height 60 m from the ground at time t = 0.
Further, `vec(a_("COM"))=(m_(A)vec(a_(A))+m_(B)vec(a_(B))/(m_(A)+m_(B))`
`=g=10m//s^(2)` (downwards) as `vec(a_(A))=vec_(a_(B))=g` (downwards)
`vec(u_("COM"))=(m_(A)vec(U_(A))+m_(B)vec(U_(B)))/(m_(A)+m_(B))`
`=((1)(200)-(2)(50))/(1+2)=100/3` m/s (upwards)
Let, h be the height attained by COM beyond 60 m. Using,
`v_("COM")^(2)=u_("COM")^(2)+2a_("COM")h`
or `0=(100/3)^(2)-(2)(10)h`
or `h=((100)^(2))/180=55.55m`
Therefore, maximum height attained by the centre of mass is
H = 60 + 55.55 = 115.55 m