Question

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at $2\frac{m}{s^2}$. He reaches the ground with a speed of $3\frac{m}{s}$. 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 $v^2=u^2+2as=0+2\times 9.8\times 50=980$
$[Asu=0,a=\frac{9.8m}{s^2},s=50m]$
At point B, parachute opens and it moves with retardation of $2\frac{m}{s^2}$ and reach at ground (Point C) with velocity of $3\frac{m}{s}$
For the part 'BC' by applying the equation $v^2=u^2+2as$
$v=\frac{3m}{s},u=\sqrt{980}\frac{m}{s},a=\frac{-2m}{s^2},s=h$
$\Rightarrow (3{)}^2=(\sqrt{980}{)}^2+2\times (-2)\times h\Rightarrow 9=980-4h$
$\Rightarrow h=\frac{980-9}{4}=\frac{971}{4}=242.7\cong 243m.$
So, the total height by which parachutist bail out = 50 + 243 = 293m.