Two particles `A` and `B` are thown vertically upward with velocity, vertically upward with velocity, `5 m s^(-1)` and `10 m s^(-1)` respectively (g=10 m s^(-2)), Find separation between them after `1 s`.
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Explanation

Method -1: Position of `A` aftre `1` s
`Sa_(A)=ut-(1)/(2) g t^(2)`
`=5t-(1)/(2) xx10xx t^(2)=5xx1-5xx1^(2)=5-5=0`
i.e., the particle will return to ground at `t=1 s`
`S_(B)=ut-(1)/(2) g t^(2)=10xx1-(1)/(2)xx10xx1^(2)=10-5=5 m`
Hence, separation between `A` and `B`, `S_(B)-S_(A)=5 m`
Method-2: Relative velocity method:
`vec a_(BA)=vec v_(A)=10-5=5 m s^(-1)`
Hence, relative separation between particles
`becuase vec s_(BA) (in 1 s)=vec v_(BA)xxt=5xx1=5 m`
becuase Distance between `A` and `B` after `1 s =5 m`.
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