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Clearly distance moved by 1st ball in 18s = distance moved by 2nd ball in 12s. Distance moved in 18 s by 1st ball = 1/2 × 10 × 18² = 1620 m. Distance moved in 12 s by 2nd ball = u + 1/2 g(12)².

1. h = 1/2 g(5)² = 125
2. h1 + h2 + h3 = 1/2 g(10)² = 500
3. h1 = h2/3 = h3/5
4. The distance covered in time t, 2t, 3t, etc. will be in the ratio of 1² : 2² : 3² i.e., square of integers i.e., h ∝ t².

Correct answer: The distance covered in time t, 2t, 3t, etc. will be in the ratio of 1² : 2² : 3² i.e., square of integers i.e., h ∝ t².

Solution

The distance covered by an object under uniform acceleration is proportional to the square of the time (h ∝ t²). This is derived from the equation of motion: s = ut + 1/2 at², where initial velocity u = 0 and acceleration a = g.

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