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Q.

A parachutist drops freely from an aeroplane for 8s before the parachute opens out. Then he descends with a net retardation of 2 ms-2 reaching the ground with a velocity of 6 ms-1. The height from which he bails out of the
aeroplane is (g = 10 ms-2)

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a

1929 m

b

1911 m

c

1591 m

d

2000 m

answer is C.

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Detailed Solution

v = u + at 
   = 0 + 10×8 
   =80 m/s 

s = ut + 1/2​at2 

  = 0 + 1/2​×10×82 

  = 320 m 

vf2 ​= vi2 ​+ 2as 

s = (vf2 ​−vi2​​)/2a 

  = (62−802)​ /(2 x -2)

  =1591 m

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