A ball is dropped from the top of a tower. After 2 s another ball is thrown vertically downwards with a speed of `40 ms^(-1)`. After how much time and at what distance below the top of tower the balls meet ?

Explanation

To solve the problem, we need to analyze the motion of both balls separately and then find the time and distance at which they meet.

### Step 1: Analyze the first ball's motion
The first ball is dropped from the top of the tower.
- Initial velocity (u1) = 0 m/s (since it is dropped)
- Acceleration (a) = g = 9.8 m/s² (acceleration due to gravity)
- Time (t1) = t seconds (we will find this later)

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