Two particles A and B of masses 1 kg and 2 kg respectively are projected in the directions shown in figure with speeds `u_(A)=200m//s and u_(B)=50m//s`. Initially they were 90 m apart. They collide in mid air and stick with each other. Find the maximum height attained by the centre of mass of the particles. Assume acceleration due to gravity to be constant. `(g=10m//s^(2))`

Options

A. 151.55 m
B. 105.55 m
C. 115.55 m
D. 125.55 m

Answer

C

Explanation

Using `m_(A)r_(A)=m_(B)r_(B)`
`r_(A)=2r_(a)" .............(1)"`
also `r_(A)+r_(B)=90" ..............(2)"`
On solving, `r_(A)=60m and r_(B)=30m`
ie, COM is at height 60 m from the ground at time t = 0.
Further, `a_(COM)=(m_(A)a_(A)+m_(B)a_(B))/(m_(A)+m_(B))=g=10m//s^(2)," "a_(A)=a_(B)=g`
`u_(COM)=(m_(A)u_(A)+m_(B)u_(B))/(m_(A)+m__(B))=100/3m//s`
Let h be the height attained by COM beyond 60 m , `" Using "v_(COM)^(2)=u_(COM)^(2)+2a_(COM)h`
`0=(100/3)^(2)-(2)(10)h, h=(100)^(2)/180=55.55m therefore` maximum height attained by COM is
`H=60+55.55=115.55m`