Question

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at $2m/s^2$ . He reaches the ground with a speed of $3m/s$ . At what height, did he bail out ? (Given $g=9.8m/s^2$ approximately)

• A. 91 m
• B. 182 m
• C. 293 m
• D. 111 m

Answer 1

Answer: (C)

293 m

The speed of the person just before the parachute opens is:
$u=\sqrt{2gh}=\sqrt{2\times 9.8\times 50}=\sqrt{980}$

When the parachute opens and descends,

$v^2=u^2+2(-a)h$

$v^2-u^2=-2\left(2\right)\left(h\right)$

$9-980=-4h$
$\Rightarrow h=971/4=242.75m$

$\therefore$ The total height of fall is $242.75+50=292.75m\approx 293m$