Question

A parachutist drops freely from an airplane for $10\, s$ before the parachute opens. He then descends. with a uniform retardation of $2.5 \,ms ^{-2}$. If he bails out of the plane at a height of $2495\, m$ and $g$ is $10\, ms ^{-2}$, his velocity on reaching the ground will be

• A. $ 5\,m{{s}^{-1}} $
• B. $ 10\,m{{s}^{-1}} $
• C. $ 15\,m{{s}^{-1}} $
• D. $ 20\,m{{s}^{-1}} $

Answer 1

Answer: (A)

Initial velocity of dropping $=$ zero
Let $v_{1}$ be velocity at end of $10 \,s$
$\Rightarrow v_{1}=g t $
$\Rightarrow v_{1}=100\, ms ^{-1}$
Distance travelled during this time is
$h_{1}=\frac{v_{1}^{2}}{2 g}=\frac{(100)^{2}}{2(10)}$
$\Rightarrow h_{1}=500 \,m$
So, a remaining distance of $2495-500=1995\, m$ has to be travelled with a retardation of $2.5\, ms ^{-2}$.
Let the parachutist strike the ground with velocity $v$.
Then $ v^{2}-v^{'2}=2 a\left(h-h^{'}\right)$
$\Rightarrow v^{2}-(100)^{2}=2(-2.5)(1995) $
$\Rightarrow v^{2}=10000-9975$
$\Rightarrow v^{2}=25 $
$\Rightarrow v=5\, ms ^{-1}$