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PhysicsNEET 2021Projectile motion

Motion in a Plane

A particle moving in a circle of radius RR with a uniform speed takes a time TT to complete one revolution. If this particle were projected with the same speed at an angle θ\theta to the horizontal, the maximum height attained by it equals 4R4R. The angle of projection θ\theta is then given by:

  1. A.θ=cos⁡−1(gT2π2R)1/2\theta = \cos^{-1}\left(\dfrac{gT^2}{\pi^2R}\right)^{1/2}
  2. B.θ=cos⁡−1(π2RgT2)1/2\theta = \cos^{-1}\left(\dfrac{\pi^2R}{gT^2}\right)^{1/2}
  3. C.θ=sin⁡−1(π2RgT2)1/2\theta = \sin^{-1}\left(\dfrac{\pi^2R}{gT^2}\right)^{1/2}
  4. D.θ=sin⁡−1(2gT2π2R)1/2\theta = \sin^{-1}\left(\dfrac{2gT^2}{\pi^2R}\right)^{1/2}✓

You have the answer. The worked solution — and why the other three were written — is part of a paid pack.

More from this chapter in past papers

From Physics › Motion in a Plane › Projectile motion — lectures, notes and every past question on it.

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