Question

Questions 1. A parachutists bails out from an aeroplane and after dropping through a distance of 40m opens the parachute and decelerates at 2 m/s sq. If he reaches the ground with a speed of 2m/s, how long is he in the air? At what height did he bail out from the plane.

Answer 1

After 40m his speed is,
$v^2 = 2gs$
$v^2= 2*9.81*40$
$v= 28.01$
now after the opening of parachute,
acceleration,a = -2
final velocity, v= 2
initial velocity, u = 28.01
distance traveled, s =?
$v^2= u^2 + 2 a s$
$s= \frac{v^2 - u^2}{2a}$
$s=\frac{2^2-28.01^2}{2*(-2)}$
$s=195.14$
so,
height$=195.14+40=235.14$