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3.A particle is thrown vertically up with initialvelocity of 60 m/s. The distance covered by theparticle in first two seconds of descent will be(take g = 10 m/s2)(1) 5 m(2) 15 m(3) 20 m(4) 40 mAakash Educational Services Limited - Regd. Office : Aakash |
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Answer» answer : OPTION (3) 20m this question is little different. question want to find distance covered by particle in first two SECONDS of descent. particle first moves upward, after reaching a specific height, it falls DOWNWARD (descending motion). here observing point is that at specific height velocity of particle becomes zero after then it will ready to FALL. so, initial velocity of particle in descending motion, u = 0 distance travelled by particle , s = ut + 1/2 at² here, u = 0, t = 2s and a = -g = -10m/s² then, s = 0 + 1/2 (-10m/s²) × (2s)² = -20m [ here negative sign indicates that particle is moving downward ] hence, distance covered by particle in first two seconds of descent will be 20m |
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