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

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1929 m

1911 m

2195 m

2000 m

answer is B.

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Detailed Solution

AC = AB + BC

$AB = \frac{1}{2} {gt}^{2} = \frac{1}{2} \times 10 \times 64 = 64 \times 5 = 320$

$v^{2} - u^{2} = 2 as$

$6^{2} - {(80)}^{2} = 2 (- 2) s$

$36 - 6400 = - 4s$

$6364 = 4s$

$s = \frac{6364}{4} = 1591$

$AC = AB + AC = 320 + 591 = 1911 m$

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