Q.A parachutist bails out from an aero plane and after dropping through a distance of 40m, he opens his parachute and decelerates at 2m/s. If he reaches the ground with a speed of 2m/s, how long is in air? At what height did he bail out of the plane?

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Solution

Initial velocity u = 0
Travels 40 m in time t with acceleration g. $s = u t + \frac{1}{2} a t^{2} 40 = 0 + \frac{1}{2} \times 10 \times t^{2} t = 2.82 s v e l o c i t y a f t e r t i m e t v = u + g t v = 0 + 10 \times 2.82 v = 28.2 m / s W h e n p r a c h u t i s t o e n s t h e p a r a c h u t e, i n t i a l v e l o c i t y i s 28.2 m / s d e a c c e l e r a t e s a t 2 m / s^{2} a n d r e a c h t o t h e g r o u n d w i t h 2 m / s . h e n c e f i n a l v e l o c i t y i s 2 m / s 2 = 28.2 - 2 t_{1} w h e r e t_{1} i s t h e t i m e t a k e n t o r e a c h t h e g o u n d . t_{1} = 13.1 s d i s \tan c e t r a v e l i n t_{1} s s_{1} = u t_{1} + \frac{1}{2} a {t_{1}}^{2} = 28.2 \times 13.1 - \frac{1}{2} \times 2 \times 13.1 \times 13.1 = 197.81 m t o t a l t i m e t a k e n = 13.1 + 2.82 = 15.92 s h e i g h t f r o m g r o u n d = 197.81 + 40 = 237.81 m$

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