Question

A parachutist drops first freely from an airplane for $10s$ and then the parachute opens out. Now he descends with 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

• A. $5\text{m/s}$
• B. $10\text{m/s}$
• C. $15\text{m/s}$
• D. $20\text{m/s}$

Answer 1

The correct option is A $5\text{m/s}$

The velocity $v$ acquired by the parachutist after $10s$,
$v=u+gt$ (By first equation of motion)
$=0+10\times 10$
$=100\text{m/s}$
Distance covered by the parachutist during $10s$ of falling freely is given by $s_1=ut+\frac{1}{2}g{t}^2$ (By second equation of motion)
$s_1=0+\frac{1}{2}\times 10\times {10}^2=500\text{m}$
The distance travelled by the parachutist under retardation,
$s_2=2495-500=1995\text{m}$
Let $v_g$ be this velocity on reaching the ground.
Then, $v_g^2-v^2=2a{s}_2$ (By third equation of motion)
$v_g^2-(100{)}^2=2\times (-2.5)\times 1995$
$v_g^2=-(5\times 1995)+{100}^2=25$
$v_g=5\text{m/s}$