A parachutist after bailing out falls freely for 50 m. Then his parachute opens and he falls with deceleration of $2 m s^{- 2}$ point when he reaches the ground his speed is reduced to only 3 meters per second. At what height did he bailout? For how long was he in air?

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Solution

$S = u t + (\frac{1}{2}) a t^{2} 50 = 0 + (\frac{1}{2}) \times 10 \times t_{1}^{2} ∴ t_{1}^{2} = 10 ∴ t_{1} = \sqrt{10} s e c a l l v^{2} = u^{2} + 2 a s v^{2} = 0 + 2 \times 10 \times 50 ∴ v^{2} = 1000 n o w w h e n p a r a c h u t e o p e n s, i t s t a r t s d e a c c e l e r a t i n g a n d r e a c h e s g r o u n d w i t h v e l o c i t y 3 m / s e c ∴ v^{2} = u^{2} + 2 a s 3^{2} = 1000 + 2 \times (- 10) \times 5 5 = (\frac{1000 - 9}{20}) = 49.55 m h e n c e h e i g h t a t w h i c h h e b a i l s o u t = 50 + 49.55 = 99.55 m a n d V = u + a t o r 3 = \sqrt{1000} - 10 \times t_{2} ∴ t_{2} = (\frac{10 \sqrt{10} - 3}{10}) = 2.86 s e c ∴ t o t a l t i m e = t_{1} + t_{2} = \sqrt{10} + 2.86 = 6 s e c (a p p r o x)$

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