Question

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

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

Answer 1

Answer: (A)

Initial velocity of dropping $=$ zero
Let $v_1$ be velocity at end of $10s$
$\Rightarrow v_1=gt$
$\Rightarrow v_1=100m{s}^{-1}$
Distance travelled during this time is
$h_1=\frac{v_1^2}{2g}=\frac{(100{)}^2}{2(10)}$
$\Rightarrow h_1=500m$
So, a remaining distance of $2495-500=1995m$ has to be travelled with a retardation of $2.5m{s}^{-2}$.
Let the parachutist strike the ground with velocity $v$.
Then $v^2-v^{{}^{\prime }2}=2a\left(h-h^{{}^{\prime }}\right)$
$\Rightarrow v^2-(100{)}^2=2(-2.5)(1995)$
$\Rightarrow v^2=10000-9975$
$\Rightarrow v^2=25$
$\Rightarrow v=5m{s}^{-1}$