A ball is throuwn with a speed of 20 m//s from top of a buling 150 m high and simultaneously another ball is thrown vertically upward witn a speed of 30 m/s from the foot of the builing . Find the time when both the balls will meet `(g =10 m//s^2)`

Explanation

`(I) S_1 = 20 t+5t^2 = 30 t- 5 t^2`
`150=50 t `
` implies t= 3s `

(ii) Relative acceleration of both is zero since both have acceleration in downward direction .
` veca _(AB) = veca_(A) - veca_B = g-g=0`
` vecV _(BA) = 30 -(-20) = 50`
` S_(nA) = V_(BA) xx t `
` t=(S_(BA))/(V_(BA)) =( 150 )/(50) - 3S `