Question

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

Answer 1

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