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…

Question

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)

Accepted Answer

Correct answer: D. (d) During the first 2 s of free fall, the paratrooper starts from rest ( u=0 ). Velocity after 2 s is v_1=u+ gt=0+10 2=20 m/s. Distance covered during free fall is h_1= 12gt^2= 12 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 v_2=5 m/s. Let h_2 be the distance covered during deceleration until reaching 10 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 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 m.

Solution

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