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A ball is thrown up with a certain velocity so that it reaches a height 'h'. Find the ratio of the two different times of the ball reaching h/3 in both the directions.

1. (√2 − 1) / (√2 + 1)
2. 1 / 3
3. (√3 − √2) / (√3 + √2)
4. (√3 − 1) / (√3 + 1)

Correct answer: (√3 − √2) / (√3 + √2)

Solution

With u=sqrt(2gh), the two times to reach h/3 satisfy g t^2-2u t+2h/3=0, giving t=[sqrt(2gh)+/-sqrt(4gh/3)]/g. The ratio of smaller to larger root simplifies to (sqrt3-sqrt2)/(sqrt3+sqrt2) (approx 0.101).

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