Question

A parachutist after bailing out falls $50\text{m}$ without friction. When parachute opens, it decelerates at $2{\text{m/s}}^2$. He reaches the ground with a speed of $3\text{m/s}$. 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 at the end of free fall is ${}^{\prime }{v}^{\prime }$.

From, $v^2=u^2+2as=0+2\times 9.8\times 50=980$
[As $u=0$, $a=9.8{\text{m/s}}^2$, $s=50\text{m}$]
So, $v^2=980$

At point B, parachute opens, and it moves with retardation of $2{\text{m/s}}^2$ and reach at ground (Point C) with velocity of $3\text{m/s}$.

For the part 'BC' by applying the equation we get:
$v^2=u^2+2as$
$v=3\text{m/s},u=\sqrt{980},a=-2{\text{m/s}}^2,s=h$

$\Rightarrow (3{)}^2=(\sqrt{980}{)}^2+2\times (-2)\times h\Rightarrow 9=980-4h$

$\Rightarrow h=\frac{980-9}{4}=\frac{971}{4}=242.7\cong 243m$

So, the total height by which parachutist bail out $=50+243=293\text{m}$