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. Then

Options

A. Both A and B will reach the ground simultaneously
B. 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 velocity
D. Both the balls will hit the ground with the different velocity.

Answer

A, C, D

Explanation

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.