Question

A parachutist after bailing out falls 50 m without friction. When parachut iling 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] (A) 91 m (B) 182 m (C)293 m (D) 111m verti (A)

JEE/Engineering Exams

Physics

Solution

114

4.0 (1 ratings)

Let The Height Be "s" m. Distance Covered Without Parachute ( =50 mathrm{m} ) Initial Velocity ( =0 ) Acceleration ( =9.8 mathrm{m} / mathrm{sec} 2 ) ( v^{2}=u^{2}+2 a s ) ( v^{2}=2^{star} 9.8^{*} 50=980 ) ( v^{2}=980 ) ( v=operatorname{root} 980=14^{star} operatorname{Root} 5 mathrm{m} / mathrm{sec} ) Now Remaining Distance ( =s-50 ) Initial Velocity ( =14^{star} ) Root ( 5 mathrm{m} / mathrm{sec} ) Acceleration ( =-2 mathrm{m} / mathrm{sec}^{2} ) ( v^{2}=u^{2}+2 a s ) ( 3^{2}=980-4(s-50) ) ( 9=980-4 s+200 ) ( 9=1180-4 s ) ( 4 s=1171 ) ( s=292.75 mathrm{m} )