Question

A man bails out form an aeroplane and after dropping vertically through a distance of 40 m, he opens the parachute and decelerates at 2$m{s}^{-2}.$ If he reaches the ground with a speed of 2 𝑚/𝑠, how much distance did he travel in air ?

Answer 1

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