Q. A parachutist after bailing out falls $50\, m$ without friction. When parachute opens, it decelerates at $2\,m/s^2$. He reaches the ground with a speed of $3 \,m/s$. At what height, did he bail out ?
AIEEEAIEEE 2005Motion in a Straight Line
Solution:
Parachute bails out at height $H$ from ground.Velocity at $A$
$v=\sqrt{2gh}$
$=\sqrt{2\times9.8\times50}$
$=\sqrt{980}\,m/s$
The velocity at ground $v_{1}=3\,m/s$ (given)
Acceleration $= - 2 m/s^2$ (given)
$\therefore H-h=\frac{v^{2}-v^{2}_{1}}{2\times2}$
$=\frac{980-9}{4}$
$=\frac{971}{4}=242.75$
$\therefore H=242.75+ h$
$=242.75+50\approx293\,m$
