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Acceleration

A parachutist...

Question

A parachutist drops freely from an aeroplane for 10 s before the parachute opens out.Then he descends with net retardation 2.5.If the bails out of the plane at a height of 2495m,his velocity on reaching ground will be?

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Solution

$U \sin g v = u + g t u = 0, v = - 10 \times 10 = - 100 N o w u \sin g s = u t + 1 / 2 u t^{2} s = 0 + \frac{1}{2} \times - 10 \times 100 s = 500 H = 2495 - 500 = 1995 v^{2} = u^{2} + 2 a h v^{2} = 10000 - 2 \times 2.5 \times 1995 = 25 v = 5 m / s e c$

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