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A particle is projected straight upward with speed u. During its ascent, it crosses three points A, B, and C with speeds u/2, u/3, and u/4 respectively. The ratio of the distances AB and BC is

1. 20: 7
2. 2
3. 10: 7
4. 1

Correct answer: 20: 7

Solution

Height for speed v is h=(u^2-v^2)/2g. With v = u/2, u/3, u/4 the heights are (3/4, 8/9, 15/16)(u^2/2g). AB = 8/9-3/4 = 5/36 and BC = 15/16-8/9 = 7/144 (in units of u^2/2g), giving AB:BC = (20/144):(7/144) = 20:7.

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