Q162 / 180 · KCET 2007 · Physics
The maximum height attained by a projectile when thrown at an angle $\theta$ with the horizontal is found to be half the horizontal range. Then $\theta$ is equal to
- Correct answerA$\tan ^{-1}$ (2)
- B$\frac{\pi}{6}$
- C$\frac{\pi}{4}$
- D$\tan ^{-1}\left(\frac{1}{2}\right)$
Solution
Maximum height, $H_{0}=\frac{u^{2} \sin _{2} \theta}{2 g}$
Range, $\quad R=\frac{u^{2} \sin 2 \theta}{g}$
Given, $\quad H_{0}=\frac{R}{2}$
$$
\begin{array}{llrl}
& \therefore & \frac{u^{2} \sin ^{2} \theta}{2 g} & =\frac{u^{2} 2 \sin \theta \cos \theta}{2 g} \\
\Rightarrow & & \sin \theta & =2 \cos \theta \\
\Rightarrow & & \tan \theta & =2 \\
& \therefore & \theta & =\tan ^{-1}(2)
\end{array}
$$
Hence, the correct answer is option: 'A'.