JEE Main 2026 — Motion In One Dimension Question with Solution

During the first $2\text{ s}$ of free fall, the paratrooper starts from rest ($u=0$).
Velocity after $2\text{ s}$ is $v_1=u+gt=0+10\times 2=20\text{ m/s}$.
Distance covered during free fall is $h_1=\frac{1}{2}g{t}^2=\frac{1}{2}\times 10\times 2^2=20\text{ m}$.
After opening the parachute, the paratrooper decelerates at $a=-3{\text{ m/s}}^2$.
At a height of $10\text{ m}$ from the ground, the final velocity is $v_2=5\text{ m/s}$.
Let $h_2$ be the distance covered during deceleration until reaching $10\text{ m}$ height.
Using $v_2^2=v_1^2+2a{h}_2$:
$5^2=20^2+2(-3)h_2$
$25=400-6h_2$
$6h_2=375$
$h_2=62.5\text{ m}$.
The total height of the airplane is the sum of the free fall distance, the deceleration distance, and the remaining height from the ground.
$H=h_1+h_2+10=20+62.5+10=92.5\text{ m}$.