A Parachutist bails out from an aeroplaneand after dropping through a distance of 40m opens the parachute and decelerates at 2m/s2 if he reach the ground with a speed of 2m/s how long he is in air? at what height did he bail out from the plane

Travels 40 m in time t with acceleration g.

$s=ut+\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}40=0+\frac{1}{2}\times 10\times {t}^{2}\phantom{\rule{0ex}{0ex}}t=2.82s\phantom{\rule{0ex}{0ex}}velocityaftertimet\phantom{\rule{0ex}{0ex}}v=u+gt\phantom{\rule{0ex}{0ex}}v=0+10\times 2.82\phantom{\rule{0ex}{0ex}}v=28.2m/s\phantom{\rule{0ex}{0ex}}Whenprachutistoenstheparachute,intialvelocityis28.2m/s\phantom{\rule{0ex}{0ex}}deacceleratesat2m/{s}^{2}andreachtothegroundwith2m/s.hencefinalvelocityis2m/s\phantom{\rule{0ex}{0ex}}2=28.2-2{t}_{1}\phantom{\rule{0ex}{0ex}}where{t}_{1}isthetimetakentoreachthegound.\phantom{\rule{0ex}{0ex}}{t}_{1}=13.1s\phantom{\rule{0ex}{0ex}}dis\mathrm{tan}cetravelin{t}_{1}s\phantom{\rule{0ex}{0ex}}{s}_{1}=u{t}_{1}+\frac{1}{2}a{{t}_{1}}^{2}\phantom{\rule{0ex}{0ex}}=28.2\times 13.1-\frac{1}{2}\times 2\times 13.1\times 13.1\phantom{\rule{0ex}{0ex}}=197.81m\phantom{\rule{0ex}{0ex}}totaltimetaken=13.1+2.82=15.92s\phantom{\rule{0ex}{0ex}}heightfromground=197.81+40=237.81m\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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