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

• A

  $60^{\circ}$
• Correct answer
  B

  $\tan ^{-1}\left(\frac{1}{2}\right)$
• C

  $\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• D

  $45^{\circ}$

Solution

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 $\alpha$. $\therefore \text { In } \triangle P O M \text {, }$ $\tan \alpha=\frac{H}{R / 2}=\frac{\frac{u^2 \sin ^2 \theta}{2 g}}{\frac{u^2 \sin 2 \theta}{2 g}}=\frac{\sin ^2 \theta}{\sin 2 \theta}=\frac{\sin ^2 45^{\circ}}{\sin 90^{\circ}}$ $=\frac{\left(\frac{1}{\sqrt{2}}\right)^2}{1} \Rightarrow \tan \alpha=\frac{1}{2}$ $\Rightarrow \quad \alpha=\tan ^{-1}\left(\frac{1}{2}\right)$

Hence, the correct answer is option: 'B'.