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JEE Main 2021 (31 Aug, Shift 1) · Horizontal Projectile & Inclined Plane

Horizontal Projectile & Inclined Plane question from JEE Main 2021 (31 Aug, Shift 1)

Q7JEE Main 2021 · 31 AUG · SHIFT 1Easy

Horizontal Projectile & Inclined Plane

A helicopter is flying horizontally with a speed 'v' at an altitude 'h' has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped?

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Answer: (c)

  1. Time of fall t=2hgt = \sqrt{\frac{2h}{g}}; in this time the packet moves x=vt=v2hgx = vt = v\sqrt{\frac{2h}{g}} horizontally.
  2. The man is at horizontal distance xx and vertical distance hh from the helicopter.
  3. Distance =x2+h2=2v2hg+h2= \sqrt{x^2 + h^2} = \sqrt{\frac{2v^2h}{g} + h^2}
  4. So the answer is (c).
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