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A parachutist drops freely from an aero plane for 10 s before the parachute opens out. Then he descends with a net retardation of $2.5 m s^{- 2}$ . If he bails out of the plane at a height of 2495 m and $g = 10 m s^{- 2}$ , his velocity on reaching the ground will be
(in $m s^{- 1}$ )

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answer is 5.

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Detailed Solution

Suppose the man drops at A, from A to B he is falling freely & then at B parachute opens out & he falls with a retardation of .

$∴ A B = \frac{1}{2} \times 10 \times 10^{2} = 500 m$

$∴ B C = A C - A B = 2495 - 500 = 1995 m$

Velocity at B,

$V_{B} = g t = 10 \times 10 = 100 m / s ↓$

Velocity at C,

$V_{C} = \sqrt{{V_{B}}^{2} + 2 a y} = \sqrt{100^{2} + 2 \times 2.5 \times (- 1995)}$

$= \sqrt{25} = 5 m / s ↓$

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