Kinematics

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Question 40 of 308

A paratrooper jumps from an aeroplane and opens a parachute after 2 s of free fall and starts deaccelerating with 3 m/s23 \mathrm{~m} / \mathrm{s}^{2}. At 10 m height from ground, while descending with the help of parachute, the speed of paratrooper is 5 m/s5 \mathrm{~m} / \mathrm{s}. The initial height of the airplane is ____\_\_\_\_ m.
(g=10 m/s2)\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)

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Answer: (D) 92.5

During the first 2 s2\text{ s} of free fall, the paratrooper starts from rest (u=0u = 0).
Velocity after 2 s2\text{ s} is v1=u+gt=0+10×2=20 m/sv_1 = u + gt = 0 + 10 \times 2 = 20\text{ m/s}.
Distance covered during free fall is h1=12gt2=12×10×22=20 mh_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 m/s2a = -3\text{ m/s}^2.
At a height of 10 m10\text{ m} from the ground, the final velocity is v2=5 m/sv_2 = 5\text{ m/s}.
Let h2h_2 be the distance covered during deceleration until reaching 10 m10\text{ m} height.
Using v22=v12+2ah2v_2^2 = v_1^2 + 2ah_2:
52=202+2(−3)h25^2 = 20^2 + 2(-3)h_2
25=400−6h225 = 400 - 6h_2
6h2=3756h_2 = 375
h2=62.5 mh_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=h1+h2+10=20+62.5+10=92.5 mH = h_1 + h_2 + 10 = 20 + 62.5 + 10 = 92.5\text{ m}.

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