1.

A stone A is dropped from rest from a height h above the ground . A second stone B is simultaneously thrown verticallly up from a point on the ground with velocity u. The line of motion of both the stone is same . Find if the value of v which would enable the stone B to meet the stone A midway between their initial positions Time of travel of each stone =

Answer»

Solution :DISTANCE TRAVELLED by each stone `=h/2`
For stone `A, h/2=1/2"GT"^(2)`i.e.,`t=sqrt(h/g)`
For stone `B,h/2=ut-1/2"gt"^(2)=usqrt(h/g)-1/2g(h/g)`
`rArrh/2=usqrt(h/g)-h/2"(or)"usqrt(h/g)=h:. U=hsqrt(h/g)=sqrt(GH)`


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