During the first 2 s of free fall, the paratrooper starts from rest (u=0).
Velocity after 2 s is v1=u+gt=0+10×2=20 m/s.
Distance covered during free fall is h1=21gt2=21×10×22=20 m.
After opening the parachute, the paratrooper decelerates at a=−3 m/s2.
At a height of 10 m from the ground, the final velocity is v2=5 m/s.
Let h2 be the distance covered during deceleration until reaching 10 m height.
Using v22=v12+2ah2:
52=202+2(−3)h2
25=400−6h2
6h2=375
h2=62.5 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=h1+h2+10=20+62.5+10=92.5 m.