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NEET Physics QUESTION #7339
Question 1

A projectile is fired at $45°$ to the horizontal. The elevation angle of the projectile at its highest point, as seen from the point of projection, is:

  • $45°$
  • $60°$
  • $\tan^{-1}\!\left(\dfrac{1}{2}\right)$✔️
  • $\tan^{-1}\!\left(\dfrac{\sqrt{3}}{2}\right)$
Correct Answer Explanation

At max height $H = \dfrac{u^2}{4g}$; horizontal distance $= \dfrac{R}{2} = \dfrac{u^2}{2g}$

Elevation angle: $\tan\phi = \dfrac{H}{R/2} = \dfrac{u^2/(4g)}{u^2/(2g)} = \dfrac{1}{2}$

$\phi = \mathbf{\tan^{-1}\!\left(\dfrac{1}{2}\right)}$