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  4. A bullet fired from a gun falls at a distance half of its maximum range. The angle of projection of the bullet is

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Q. A bullet fired from a gun falls at a distance half of its maximum range. The angle of projection of the bullet is

AP EAMCETAP EAMCET 2019

Solution:

As, we know that for a projectile motion, maximum range,
$R_{\max }=\frac{u^{2}}{g}$ and $h_{\max }=\frac{R_{\max }}{4}$
Given, $ R=\frac{R_{\max }}{2}$
$\Rightarrow \, \frac{R_{\max }}{2}=\frac{u^{2}}{2 g}=R\,\,\,\,\,\dots(i)$
where, $R=\frac{u^{2} \sin 2 \theta}{g}$
Hence, $ \frac{u^{2} \sin 2 \theta}{g}=\frac{u^{2}}{2 g}$ [From Eq. (i)]
$\Rightarrow \, \sin \,2 \theta=\frac{1}{2}$
$\Rightarrow \,2 \theta=30^{\circ} $
$\Rightarrow \, \theta=15^{\circ}$