- Home
- Standard 11
- Physics
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 ?........$m$
$293$
$111$
$91$
$182$
Solution
(a) 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 $u = 0$, $a = 9.8m/{s^2}$, $s = 50\, m$]
At point $B$, parachute opens and it moves with retardation of 2$m/{s^2}$ and
reach at ground (Point $C$) with velocity of $3\,m/s$
For the part $‘BC’$ by applying the equation ${v^2} = {u^2} + 2as$
$v = 3\,m/s$, $u = \sqrt {980} \,m/s$, $a = – 2m/{s^2}$, $s = h$
$⇒ {(3)^2} = {(\sqrt {980} )^2} + 2 \times ( – 2)\, \times \,h ⇒ 9 = 980 – 4h$
$⇒ h = \frac{{980 – 9}}{4}$ $ = \frac{{971}}{4} = 242.7 \tilde = 243\,m$.
So, the total height by which parachutist bail out = $50 + 243 = 293\, m$.
