Kinematics Question 250
Question: A parachut it’st after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height, did he bail out? [AIEEE 2005]
Options:
A) 293 m
B) 111 m
C) 91 m
D) 182 m
Show Answer
Answer:
Correct Answer: A
Solution:
After bailing out from point A parachut it’st falls freely under gravity. The velocity acquired by it will v From $ v^{2}=u^{2}+2as $
$ =0+2\times 9.8\times 50 $ = 980 [As u = 0, $ a=9.8m/s^{2} $ , s = 50 m] At point B, parachute opens and it moves with retardation of 2 $ m/s^{2} $ and reach at ground (Point C) with velocity of $ 3m/s $ For the part ?BC? by applying the equation $ v^{2}=u^{2}+2as $
$ v=3m/s $ , $ u=\sqrt{980}m/s $ , $ a=-2m/s^{2} $ , s = h
therefore $ {{(3)}^{2}}={{(\sqrt{980})}^{2}}+2\times (-2)\times h $
therefore $ 9=980-4h $
therefore $ h=\frac{980-9}{4} $
$ =\frac{971}{4}=242.7\tilde{=}243 $ m. So, the total height by which parachut it’st bail out = $ 50+243 $ = 293 m.
BETA
