Question

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

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

Answer 1

The correct option is A 293 m

After bailing out from point A parachutist falls freely under gravity. The velocity acquired by it
will 'v'


From $v^2=u^2+2as=0+2\times \times 9.8\times$50=980
[As u=0,a=9.8m/s^2, s=50 m]
At point B,parachute opens and it moves with retardation of $2m/s^2$ and
reach at ground (Point C) with velocity of 3m/s

For the part 'BC' by applying the euation $v^2=u^2$+2as
v=3m/s,u=$\sqrt{980}m/s,a=-2m/s^2$, s=h
$\Rightarrow (3{)}^2=(\sqrt{980}{)}^2+2\times (-2)\times h\Rightarrow 9=980-4h$
$\Rightarrow h=\frac{980-9}{4}=\frac{971}{4}=242.7\underset{\equiv }{\sim }243m.$
So,the total height by which parachutist bail out =50+243=293m.