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NEET 2011 (AIPMT, Mains) · Projectile Basics & Trajectory

Projectile Basics & Trajectory question from NEET 2011 (AIPMT, Mains)

Q9NEET 2011 · AIPMT · MAINSMedium

Projectile Basics & Trajectory

A projectile is fired at an angle of 45o with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection, is

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Answer: (c)

  1. Highest point is at (R2,H)\left(\frac R2, H\right), so tan⁡α=HR/2=2HR\tan\alpha = \frac{H}{R/2} = \frac{2H}{R}
  2. HR=tan⁡θ4⇒tan⁡α=tan⁡θ2=12\frac HR = \frac{\tan\theta}{4} \Rightarrow \tan\alpha = \frac{\tan\theta}{2} = \frac12
  3. So the answer is (c) tan⁡−1(12)\tan^{-1}\left(\frac12\right).
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