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A parachutist drops freely from an aeroplane for 10 seconds before the parachute opens out. Then he descends with a net retardation of $2 m / s^{2}$ . His velocity when he reaches the ground is $8 m / s$ . Find the height at which he gets out of the aeroplane?

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$490 m$

$98 m$

$2385 m$

$2875 m$

answer is C.

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Detailed Solution

Distance he falls before parachute open is

$S_{1} = \frac{1}{2} \times g \times \overset{50}{100} = 490 m V = u + gt = 0 + 9.8 \times 10 = 98 m / s V^{2} - u^{2} = 2 {as}_{2} {(8)}^{2} - {(98)}^{2} = 2 (- 2) S_{2} S_{2} = 2385 Total distance S = 2385 + 490 = 2875 m$

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