MCQ
A projectile is fired from level ground at an angle $\theta $ above the horizontal. The elevation angle $\phi $ of the highest point as seen from the launch point is related to $\theta $ by the relation
  • A
    $\tan \,\phi  = \frac{1}{4}\,\tan \,\theta $
  • B
    $\tan \,\phi  = \tan \,\theta $
  • $\tan \,\phi  = \frac{1}{2}\,\tan \,\theta $
  • D
    $\tan \,\phi  = 2\,\tan \,\theta $

Answer

Correct option: C.
$\tan \,\phi  = \frac{1}{2}\,\tan \,\theta $
c
It is clear from the adjoining figure that,

$\tan \phi=\frac{\mathrm{H}}{\mathrm{R} / 2}$

$=\frac{\mathrm{u}^{2} \sin ^{2} \theta / 2 \mathrm{g}}{\mathrm{u}^{2} \sin 2 \theta / 2 \mathrm{g}}=\frac{\sin ^{2} \theta}{\sin 2 \theta}=\frac{1}{2} \tan \theta$

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