A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3 ~m / s ² . At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is 5 ~m / s . The initial height of the airplane is _ _ _ _ m. ( g =10 ~m / s ² )

Step-by-step solution

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 = 1 2 gt^2 = 1 2 10 2^2 = 20 m . After opening the parachute, the paratrooper decelerates at a = -3 m/s ^2 . 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 v₂^2 = v₁^2 + 2ah₂ : 5^2 = 20^2 + 2(-3)h₂ 25 = 400 - 6h₂ 6h₂ = 375 h₂ = 62.5 m