A helicopter flying horizontally with a speed of 360 km/h at an altitude of 2 km, drops an object at an instant. The object hits the ground at a point O, 20 s after it is dropped. Displacement of 'O' from the position of helicopter where the object was released is : (use acceleration due to gravity g = 10 m/s2 and neglect air resistance)

Motion in A Plane PYQ — JEE Main 2025 · 7 Apr · Shift 2 · Q53 with Solution

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Correct answer: C

$360\,\text{km/h} = 360 \times \frac{1000}{3600} = 100\,\text{m/s}$

Since air resistance is neglected, horizontal velocity remains constant:

$x = v_x t = 100 \times 20 = 2000\,\text{m} = 2\,\text{km}$

The object falls from the helicopter to the ground, so vertical displacement is simply the altitude:

$y = 2\,\text{km} = 2000\,\text{m}$

(Indeed, using $s = ut + \tfrac12 gt^2$ with $u_y=0$ gives $s = \tfrac12 \cdot 10 \cdot 20^2 = 2000\,\text{m}$ , consistent.)

The displacement is the straight-line distance:

$R = \sqrt{x^2 + y^2}$

$R = \sqrt{(2000)^2 + (2000)^2}$

$R = 2000\sqrt{2}\,\text{m} = 2\sqrt{2}\,\text{km}$

$2\sqrt{2}\,\text{km}$

So the correct option is C.