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3-2.Motion in Plane
medium
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 $
C
$\tan \,\phi = \frac{1}{2}\,\tan \,\theta $
D
$\tan \,\phi = 2\,\tan \,\theta $
Solution
It is clear from the adjoining figure that,
$\tan \phi=\frac{\mathrm{H}}{\mathrm{R} / 2}$
$=\frac{\mathrm{u}^{2} \sin ^{2} \theta / 2 \mathrm{g}}{\mathrm{u}^{2} \sin 2 \theta / 2 \mathrm{g}}=\frac{\sin ^{2} \theta}{\sin 2 \theta}=\frac{1}{2} \tan \theta$
Standard 11
Physics
