JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Kinematics

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A helicopter rises from rest on the ground vertically upwards with a constant acceleration g . A food packet is dropped from the helicopter when it is at a height h . The time taken by the packet to reach the ground is close to [

g

is the acceleration due to gravity]:

(A) $t = 1.8 \frac{h}{g}$
(B) $t = \frac{2 h}{3 g}$
(C) $t = 3.4 (\frac{h}{g})$ ✓ Correct
(D) $t = \frac{2}{3} (\frac{h}{g})$

Answer: (C)

Solution

Rising with acceleration

g

, the helicopter reaches speed

v = 2 g h

by the time it is at height

h

, and the packet is released with this upward velocity. Taking downward as positive for the fall,

h = - 2 g h t + \frac{1}{2} g t^{2}

, that is

g t^{2} - 2 2 g h t - 2 h = 0

. The discriminant is

8 g h + 8 g h = 16 g h

, so the positive root is

t = \frac{2 2 g h + 4 g h}{2 g} = (2 + 2) \frac{h}{g} \approx 3.4 \frac{h}{g}

JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Single correct option

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JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Kinematics

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