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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:
$\tan ^{-1}\left(\frac{1}{2}\right)$
Height of projectile
$$
\begin{aligned}
& H=\frac{u^2 \sin ^2 \theta}{2 g} \\
& H=\frac{u^2 \sin ^2 45^{\circ}}{2 g} \\
& H=\frac{u^2}{4 g}
\end{aligned}
$$
Range of projectile
$$
\begin{aligned}
R & =\frac{u^2 \sin 2 \theta}{g} \\
& =\frac{u^2 \sin 90^{\circ}}{g} \\
R & =\frac{u^2}{g} \\
\therefore \quad \frac{R}{2} & =\frac{u^2}{2 g} \\
\therefore \quad \tan \alpha & =\frac{H}{R / 2}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{u^2 / 4 g}{u^2 / 2 g} \\
\tan \alpha & =\frac{1}{2} \\
\alpha & =\tan ^{-1}\left(\frac{1}{2}\right)
\end{aligned}
$$
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