The horizontal range and the maximum height of a projectile are equal. The angle of projection of the…

The horizontal range and the maximum height of a projectile are equal. The angle of projection of the projectile is
  1. $\theta=\tan ^{-1}\left(\frac{1}{4}\right)$
  2. $\theta=\tan ^{-1}(4)$
  3. $\theta=\tan ^{-1}(2)$
  4. $\theta=45^{\circ}$

Solution

Given, $R=H$
Range $R=\frac{u^2(2 \sin \theta \cos \theta)}{g}$
Height $H=\frac{u^2 \sin ^2 \theta}{2 g}$
Hence, $\frac{u^2(2 \sin \theta \cos \theta)}{g}=\frac{u^2 \sin ^2 \theta}{2 g}$
$\begin{aligned}
2 \cos \theta & =\frac{\sin \theta}{2} \\
\tan \theta & =4 \\
\theta & =\tan ^{-1}(4)
\end{aligned}$

Asked in: NEET 2012 (Screening)

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