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A \( 5 ~ k g \) stone falls from a height of \( 1000 m \) and penetrates \( 2 m \) in a layer of sand. The time of penetration is

1. 14.285 s
2. 0.0285 s
3. \( 7.146 \mathrm{s} \)
4. 0.285 s

Correct answer: 0.0285 s

Solution

The stone’s speed at the sand is found from free fall: it gains a large velocity after dropping 1000 m. Then, assuming it comes to rest uniformly over 2 m, the penetration time follows from the average-velocity relation under constant deceleration. This gives a very short time, matching 0.0285 s.

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