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An object A is dropped from rest the top of a 30m high building and at the same moment another object B is projected vertically upwards with an initial speed of 15m//s from the base of the building. Mass of the object A is 2kg while mass of the object B is 4kg. Find the maximum height above the ground level attained by the centre of mass of the A and B system (take g=10m//s^(2)) |
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Answer» Solution :`m_(1)=4kg`, `m_(2)=2kg` Initially `4kg` is on the ground `:.x_(1)=0` `2kg` is on top of the building `:.x_(2)=30m` `x_(cm)=(m_(1)x_(1)+m_(2)x_(2))/(m_(1)+m_(2))` `=(0+2xx30)/(4+2)=10m` `:.` INITIAL height of CM=`10m` Initial VELOCITY of CM, `u_(cm)=(m_(1)u_(1)+m_(2)u_(2))/(m_(1)+m_(2))` `u_(cm)=(4xx15+0)/(4+2)=10`m/s upward Acceleration of `CM`, `a_(cm)=g=10m//s^(2)` downwards `:.` Maximum height ATTAINED by CM from initial position `h_(cm)=(u_(cm)^(2))/(2g)-(10^(2))/(20)-5m` `:.` Maximum height attained by CM of `4kg` and `2kg` from the ground `=10+5=15m`
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