A block of mass \(5 \text{kg}\) rests on top of the tail of a plane of inclination \({30}^°\) preceded by a spring of constant \(2 \text{N/m}\) and contraction \(1 \ \text{m}\). The coefficient of friction between the tail surface and the block is \(0.2\). As soon as the spring is released, the plane begins to move in forward direction with an acceleration of \(1 {\text{m/s}}^2\). It takes the block \(3\) seconds to reach the bottom of the tail. As soon as the block reaches the bottom of the tail, the plane stops immediately in mid air. The distance between the tail and the tip of the plane is \(2 \text{m}\). The coefficient of friction between the block and the horizontal surface of the plane is \(0.3\). The block continues to move over the horizontal surface and falls off the tip of the plane. If the plane was flying at a height of \(50 \text{m}\) above the ground, then find the distance of the final position of the block from the tip of the plane.

The figure below would help in understanding the situation.

**Details and assumptions**:-

- Neglect the air resistance.
- Assume that the block does not rebound after hitting the ground for the first time.
- Assume that the plane is of negligible width.
- Assume that the tip of the plane is flat(as shown in figure) and not conical.
- The plane travels in the direction indicated by the \(\implies\) shown in the figure.
- Take \(g\) (acceleration due to gravity) as \(10 {\text{m/s}}^2\).
- Give your answer (in metres) to the nearest integer.

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