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  • A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection isO

A projectile is fired at an angle of 45° with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is

Option: 1

\tan^{-}\left (\frac{\sqrt{3}}{2} \right )


Option: 2

45^{\degree}


Option: 3

60^{\degree}


Option: 4

\tan^{-1}\frac{1}{2}


Answers (1)

best_answer

\\ \text{Height of projectile H}=\frac{u^{2} \sin ^{2} \theta}{2 g} \\ =\frac{u^{2} \sin ^{2} 45^{\circ}}{2 g}$

\begin{aligned} &\Rightarrow H=\frac{u^{2}}{4 g} \ldots \ldots(i)\\ &\text { Range of projectile } R=\frac{u^{2} \sin 2 \theta}{g}=\frac{u^{2} \sin 90^{\circ}}{g}\\ &\Rightarrow R=\frac{u^{2}}{g}\\ &\tan \alpha=\frac{H}{R / 2}\\ &\operatorname{In} \Delta O A B=\frac{u^{2} / 4 g}{u^{2} / 2 g}\\ &\tan \alpha=\frac{1}{2} \Rightarrow \alpha=\tan ^{-1}\left(\frac{1}{2}\right) \end{aligned}

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