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Question: Answered & Verified by Expert
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
PhysicsMotion In Two DimensionsTS EAMCETTS EAMCET 2020 (10 Sep Shift 2)
Options:
  • A $60^{\circ}$
  • B $\tan ^{-1}\left(\frac{1}{2}\right)$
  • C $\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
  • D $45^{\circ}$
Solution:
2553 Upvotes Verified Answer
The correct answer is: $\tan ^{-1}\left(\frac{1}{2}\right)$
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 $\alpha$.
$\therefore \text { In } \triangle P O M \text {, }$
$\tan \alpha=\frac{H}{R / 2}=\frac{\frac{u^2 \sin ^2 \theta}{2 g}}{\frac{u^2 \sin 2 \theta}{2 g}}=\frac{\sin ^2 \theta}{\sin 2 \theta}=\frac{\sin ^2 45^{\circ}}{\sin 90^{\circ}}$
$=\frac{\left(\frac{1}{\sqrt{2}}\right)^2}{1} \Rightarrow \tan \alpha=\frac{1}{2}$
$\Rightarrow \quad \alpha=\tan ^{-1}\left(\frac{1}{2}\right)$

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