A stone A is dropped from rest from a height h above the ground.A second stone B is simultaneously thrown vertically up from a point on the ground with velocity u.The line of motion of both the stones 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=

Explanation

Distance travelled by each stone `(h)/(2)`
For stone A,`(h)/(2)=(1)/(2)"gt"^(2)` i.e.,=`sqrt((h)/(g))-(1)g((h)/(g))`
For stone `B,(h)/(2)=ut-(1)/(2)"gt"^(2)=u sqrt((h)/(g))-(1)/(2)g((h)/(g))`
`implies(h)/(2)=u sqrt((h)/(g))-(h)/(2)` (or) `u sqrt((h)/(g))`=h `therefore=h sqrt((h)/(g))-(1)/(2)g((h)/(g))`
`implies(h)/(2)=usqrt((h)/(g))-(h)/(2)` (or) `u sqrt((h)/(g))=h therefore u=h sqrt((h)/(g))=sqrt(gh)`