From the top of a tower of height 200 m, a ball A is projected up with speed `10ms^(-1)` and 2 s later, another ball B is projected vertically down with the same speed. ThenA. Both A and B will reach the ground simultaneouslyB. Ball A will hit the ground 2 s later than B hitting the ground.C. Both the balls will hit the ground with the same velocityD. Both the balls will hit the ground with the different velocity.

From the top of a tower of height 200 m, a ball A is projected up with

1.	From the top of a tower of height 200 m, a ball A is projected up with speed `10ms^(-1)` and 2 s later, another ball B is projected vertically down with the same speed. ThenA. Both A and B will reach the ground simultaneouslyB. Ball A will hit the ground 2 s later than B hitting the ground.C. Both the balls will hit the ground with the same velocityD. Both the balls will hit the ground with the different velocity.
Answer» Correct Answer - A::C::D Ball A will return to the top of tower after `T=(2u)/(g)=(2xx10)/(10)=2s` With speed of `10ms^(-1)` downward. And this time B is also projected downwards with `10ms^(-1)`. So both reach ground simultaneously also they will hit the ground with the same speed.

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