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A projectile is fired at an angle of $45^{\circ}$ with the horizontal. Elevation angle of the projectile at its highest point as seen from the point of projection is
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Verified Answer
The correct answer is:
$\tan ^{-1}\left(\frac{1}{2}\right)$
The given situation is shown in the following figure
Angle of projection, $\theta=45^{\circ}$

The elevation angle of the projectile at its highest point as seen from the point of projection is $\alpha$.
$\therefore \text { In } \triangle P O M \text {, }$
$\tan \alpha=\frac{H}{R / 2}=\frac{\frac{u^2 \sin ^2 \theta}{2 g}}{\frac{u^2 \sin 2 \theta}{2 g}}=\frac{\sin ^2 \theta}{\sin 2 \theta}=\frac{\sin ^2 45^{\circ}}{\sin 90^{\circ}}$
$=\frac{\left(\frac{1}{\sqrt{2}}\right)^2}{1} \Rightarrow \tan \alpha=\frac{1}{2}$
$\Rightarrow \quad \alpha=\tan ^{-1}\left(\frac{1}{2}\right)$
Angle of projection, $\theta=45^{\circ}$

The elevation angle of the projectile at its highest point as seen from the point of projection is $\alpha$.
$\therefore \text { In } \triangle P O M \text {, }$
$\tan \alpha=\frac{H}{R / 2}=\frac{\frac{u^2 \sin ^2 \theta}{2 g}}{\frac{u^2 \sin 2 \theta}{2 g}}=\frac{\sin ^2 \theta}{\sin 2 \theta}=\frac{\sin ^2 45^{\circ}}{\sin 90^{\circ}}$
$=\frac{\left(\frac{1}{\sqrt{2}}\right)^2}{1} \Rightarrow \tan \alpha=\frac{1}{2}$
$\Rightarrow \quad \alpha=\tan ^{-1}\left(\frac{1}{2}\right)$
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