Question

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

Answer 1

Answer: 5

The velocity $v$ acquired by the parachutist after $10s$ :
$v=u+gt=0+10\times 10=100m{s}^{-1}$
Then, $s_1=ut+\frac{1}{2}g{t}^2=0+\frac{1}{2}\times 10\times 10^2=500m$
The distance travelled by the parachutist under retardation is
$s_2=2495-500=1995m$
Let $v_g$ be the velocity on reaching the ground. Then
$v_g^2-v^2=2a{s}_2$
or $v_g^2-(100{)}^2=2\times (-2.5)\times 1995$
or $v_g=5m{s}^{-1}$