Question

A parachutist drops first freely from an aeroplane for $10 \,s$ and then his parachute opens out. Now he descends with a net retardation of $2.5 \,m \,s^{-2}$. If he bails out of the plane at a height of $2495\, m$ and $g = 10 \,m\, s^{-2}$, his velocity on reaching the ground will be____________$ms^{-1}$

Answer 1

The velocity $v$ acquired by the parachutist after $10 \,s$ :
$v=u+gt=0+10 \times 10=100\,m\,s^{-1}$
Then, $s_{1}=ut+\frac{1}{2}gt^{2}=0+\frac{1}{2}\times 10 \times 10^{2}=500\,m$
The distance travelled by the parachutist under retardation is
$s_{2}=2495-500=1995\,m$
Let $v_{g}$ be the velocity on reaching the ground. Then
$v^{2}_{g}-v^{2}=2as_{2}$
or $v^{2}_{g}-(100)^{2}=2\times (-2.5)\times 1995$
or $v_{g}=5\,m\,s^{-1}$