A projectile is fired at an angle of $45^{\circ}$ with the horizontal. Elevation angle of the projectile at…
- $60^{\circ}$
- $\tan ^{-1}\left(\frac{1}{2}\right)$
- $\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
- $45^{\circ}$
Solution
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
$\begin{aligned}
R & =\frac{u^2 \sin 2 \theta}{g} \\
& =\frac{u^2 \sin 90^{\circ}}{g} \\
R & =\frac{u^2}{g} \\
\therefore \quad \frac{R}{2} & =\frac{u^2}{2 g} \\
\therefore \quad \tan \alpha & =\frac{H}{R / 2}
\end{aligned}$
$\begin{aligned}
& =\frac{u^2 / 4 g}{u^2 / 2 g} \\
\tan \alpha & =\frac{1}{2} \\
\alpha & =\tan ^{-1}\left(\frac{1}{2}\right)
\end{aligned}$Asked in: NEET 2011 (Mains)
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