Two balls `A` and `B` of equal masses are projected upward simultaneously, one from the ground with speed `50 ms^(-1)` and other from height `40 m` above the first ball high tower with initial speed `30ms^(-1)`. Find the maximum height attained by their centre of mass.
Text Solution
Verified by Experts
The initial position of centre of mass is `y_(ic)=(mxx0+mxx40)/(2m)=20m` Initial velocity of centre of mass `u_(cm)=(mxx50+mxx30)/(2m)=40ms^(-1)` Acceleration of centre of mass `a_(cm)=-g` Using kinematics equation `v_(cm)^2=u_(cm)^2+2a(cm)H` Here `H` is the maximum height reach by centre of mass of two balls from initial level. `:. 0^(2)=40^(2)-2xx10xxH` `impliesH=1600/20=80m` Hence maximum height reach by centre of mass from ground level will be `h_(max)=(h_(cm))_("initial")+H=20+80=100m`
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