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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Aicharya · Accepted Answer

As, we know that for a projectile motion, maximum range, $R_{\max }=\frac{u^2}{g} \text { and } h_{\max }=\frac{R_{\max }}{4}$ Given, $\quad R=\frac{R_{\max }}{2}$ $\Rightarrow \quad \frac{R_{\max }}{2}=\frac{u^2}{2 g}=R$ ...(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)] $\Rightarrow \quad \sin 2 \theta=\frac{1}{2}$ $\Rightarrow 2 \theta=30^{\circ} \Rightarrow \theta=15^{\circ}$ Hence, the correct option is (d).

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

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