A paratrooper jumps from an aeroplane and opens a parachute after 2...

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 $h_1 = \frac{1}{2}gt^2 = \frac{1}{2} \times 10 \times 2^2 = 20\text{ m}$.
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 $v_2^2 = v_1^2 + 2ah_2$:
$5^2 = 20^2 + 2(-3)h_2$
$25 = 400 - 6h_2$
$6h_2 = 375$
$h_2 = 62.5\text{ m}$.
The total height of the airplane is the sum of the free fall distance, the deceleration distance, and the remaining height from the ground.
$H = h_1 + h_2 + 10 = 20 + 62.5 + 10 = 92.5\text{ m}$.