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A particle is projected vertically upward with velocity 20m/s at the same instant another particle B is dropped from a tower of height 100m time at which B meet with A. ans is not 5sec
Asked by ashutosharnold1998 | 17 Oct, 2019, 01:19: PM
answered-by-expert Expert Answer
Time for particle-A,  which is projected vertically upwards,  to reach maximum height  is given by, 
 
t = u/g = 20/9.8 = 2.04 s
 
maximum height reached bu particle-A,  h = u2 / ( 2g ) = (20×20)/(2×9.8) = 20.41 m
 
Distance travelled by particle-B in 2.04 s , S = (1/2)g t2 = (1/2)×9.8×2.04×2.04 = 20.39 m
 
since particle-B is dropped from 100 m height, when particle-A reaches maximum height,
particl-B is ~60 m away from particle-A.
 
Hence they can not meet together during their flight.
 
Time to return to ground for particle-A is 4.08 s.
 
Time to reach ground for particle-B is obtained from,  (1/2)gt2 = 100   or  t = (200/9.8)1/2 = 4.52 s
 
Hence both will meet at 4.52 s from their staring time of motion.
Answered by Thiyagarajan K | 17 Oct, 2019, 02:54: PM

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