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JEE Main 2025 (7 Apr, Shift 2) · Horizontal Projectile & Inclined Plane

Horizontal Projectile & Inclined Plane question from JEE Main 2025 (7 Apr, Shift 2)

Q1JEE Main 2025 · 7 APR · SHIFT 2Medium

Horizontal Projectile & Inclined Plane

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)

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Answer: (c)

  1. v=360 km/h=100 m/sv = 360\text{ km/h} = 100\text{ m/s}; check t=2(2000)10=20 st = \sqrt{\frac{2(2000)}{10}} = 20\text{ s}.
  2. Horizontal distance x=100×20=2000 m=2 kmx = 100 \times 20 = 2000\text{ m} = 2\text{ km}; vertical drop =2 km= 2\text{ km}.
  3. Displacement =22+22=22 km= \sqrt{2^2 + 2^2} = 2\sqrt{2}\text{ km}
  4. So the answer is (c).
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