A steel ball is dropped from the roof of a building .An observer standing in front of a window 1.25 m high notes that the ball takes 1/8 s to fall from the top to the bottom of the window .The ball reappears at the bottom of the window 2s after passing it on the way down,if the collision between the ball reappears at the bottom of the window 2s after passing it on the way down.If the collision between the ball and the ground is perfectly elastic,then find the height of the building?Take g=10 `m//s^(2)`.

Explanation

To solve the problem step by step, we will analyze the motion of the steel ball as it falls from the roof of the building, through the window, and then bounces back after hitting the ground.

### Step 1: Analyze the motion of the ball through the window
The ball takes \( \frac{1}{8} \) seconds to fall from the top to the bottom of the window, which is 1.25 m high. We can use the equations of motion to find the velocities at the top and bottom of the window.

Let:
- \( V_0 \) = velocity at the top of the window
- \( V \) = velocity at the bottom of the window
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