From points A and B, at the respective heights of 2m and 6m, two bodies are thrown simultaneouly towards each other, one is thrown horizontally with a velocity of `8"m"/s` and the other, downward at an angle `45^(@)` to the horizontal at an initial velocity `v_(0)` such that the bodies collide in flight. The horizontal distance between points A and B equal to 8m.Then find
The initial velocity `V_(0)` of the body thrown at an angle `45^(@)`
The time of flight of the bodies before colliding
The coordinate (x,y) of the point of collision (consider the botton of the tower A as origin) is

Explanation

To solve the problem step by step, we will follow the motion equations for both bodies and find the required values.

### Step 1: Understand the Problem
We have two bodies thrown from points A and B. Body A is thrown horizontally from a height of 2 m with a velocity of 8 m/s, and body B is thrown downward at a 45-degree angle from a height of 6 m with an initial velocity \( V_0 \).

### Step 2: Set Up the Coordinate System
We will set the bottom of tower A (height = 0) as the origin (0, 0). Therefore:
- The height of point A = 2 m (coordinates: (0, 2))
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