A ball is thrown downwards with a speed of `20 m s^(-1)`, from the top of a building `150 m` high and simultaneously another ball is thrown vertically upwards with a speed of `30 m s^(-1)` from the foot to the building . Find the time after which both the balls will meet. (g=10 m s^(-2))
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Explanation

Method-1: Let the first ball move distance `S_(1)` and second ball moves up a distance `S_(2)` before they meert.
Let us take downward direction as direction as positive.
Then `S_(1)=20t +5 t^(2)`
`S_(2)=30 t-5 t^(2)`
But, `S_(1)+S_(2)=150`
`rArr 150=50 t`
`rArr t=3 s`
Method-2: Let us solve this pronlem by using the method of relative velocity .
Relative acceleration of both is zero since both have same acceleration is downward direction.
`vec a_(AB)=vec a_(A)=-vec a_(B) =g-g =0`
Initial relatie velocity,
`vec v _(BA)=30-(-20)=50 m s^(-1)`.