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Q.

A parachutist drops freely from an aeroplane for 8 sec before the parachute  opens  out. Then he descends with a net retardation of 2 m/s2 reaching the ground with a velocity of 6 m/s . The height from which the bails out of the aeroplane is  g=10 m/s2

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a

1929m

b

2195m

c

1911m 

d

2000m

answer is B.

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Detailed Solution

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After 8 sec, velocity of products

  u=v=gt=10×8=80  m/s

  Distance covered by it is

  h=12gt2=12×10×64=320m

From v2u2=2as

s=v2u22a=628022×2=3664004=63644=1591m

Total height =1591+320=11911 m

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