Question

A man bails out from an aeroplane and after dropping vertically through a distance of 40 $m$, he opens parachute and decelerates at 2$m/s^2.$ If he reaches ground with speed of 2 𝑚/𝑠, how much distance did he travel in air?
($g=9.8m/s^2$)

• A. $215m$
• B. $235m$
• C. $250m$
• D. $195m$

Answer 1

The correct option is B $235m$

After falling 40 𝑚,
he attains speed $v^2=0^2+2\times 9.8\times 40,$
$v=28m/s$

When parachutist decelerates uniformly,
$u=28m/s,v=2m/s,a=-2m{s}^{-2}$

Apply, $v^2=u^2+2as,$
$2^2={28}^2-(2\times 2\times s)$
$780=4s$
We get,$s=195m$

Height at which parachutist bailed out$=195+40=235m$