Question

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 ?

• 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 $v^2=u^2+2as=0+2\times 9.8\times 50=980$
[As u=0,a=9.8 $m/s^2$,s=50 m]
At point B,parachute opens and it moves with retardation of2 $m/s^2$ and
reach at ground (Poin C)with velocity of 3m/s.
For the part 'BC' by applying the equation $v^2=u^2+2as$

$v=3m/s,u=\sqrt{980}m/s,a=-2m/s^2,s=h$
$\Rightarrow (3{)}^2=(\sqrt{980}{)}^2+2\times (-2)\times h\Rightarrow 9=980-4h$
$\Rightarrow h=\frac{980-9}{4}=\frac{971}{4}=242.7\cong 243m.$
So, the total height by which parachutist bail out=50+243=293 m.