Two particles A and B of masses 1kg and 2kg respectively are projected in the directions shown in figure with speeds `u_A=200m//s` and `u_B=50m//s`. Initially they were `90m` 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)`

Two particles A and B of masses 1kg and 2kg respectively are projected

1.	Two particles A and B of masses 1kg and 2kg respectively are projected in the directions shown in figure with speeds `u_A=200m//s` and `u_B=50m//s`. Initially they were `90m` 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)`
Answer» `vecP_("sys")=M_(T)vecV_(CM)` `(dvecP_("sys"))/(dt)=M_(T)veca_(CM)` `vecF_(ext)=M_(T)veca_(CM)` Net external force is the gravitational force `F_(ext)=M_(T)xxg` `therefore" "veca_("cm")=gdarr`(dwonwards) `(vecV_(cm))=(m_(A)vecV_(A)+m_(B)vecV_(B))/(m_(A)+m_(B))=(1xx200-2xx50)/(3)=(100)/(3)m//suarr` (upwards) initial height of centre of mass from `A`,`h_(0)=(1xx0+2xx90)/(1+2)=60m` `h_(max)=`inital height `(h_(0))+(v_(cm)^(2))/(2g)=60+(((100)/(3))^(2))/(2xx10)=115.55m`

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