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From the top of a tower, a particle is thrown vertically downwards with a velocity of `10 m//s`. The ratio of the distances, covered by it in the `3rd` and `2nd` seconds of the motion is `("Take" g = 10 m//s^2)`.A. `5 : 7`B. `7 : 5`C. `3 : 6`D. `6 : 3` |
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Answer» Correct Answer - B `S_(3rd) = 10 + (10)/(2)(2 xx 3 -1) = 35 m` `S_(2nd) = 10 + (10)/(2) (2 xx 2 - 1) = 25 m rArr (S_(3rd))/(S_(2nd)) = (7)/(5)`. |
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