Q105 / 150 · MHT CET 2024 (04 May Shift 1) · Physics

For a projectile, the maximum height and horizontal range are same. The angle of projection ' $\theta$ ' of the projectile is

  • A
    $\tan ^{-1}\left(\frac{1}{2}\right)$
  • B
    $\tan ^{-1}(2)$
  • C
    $\tan ^{-1}\left(\frac{1}{4}\right)$
  • Correct answer
    D
    $\tan ^{-1}(4)$

Solution

$\begin{aligned} & \text { Horizontal range }=\text { Maximum Height } \\ & \frac{2 \mathrm{u}^2 \sin \theta \cos \theta}{\mathrm{~g}}=\frac{\mathrm{u}^2 \sin ^2 \theta}{2 \mathrm{~g}} \\ \therefore \quad & 2 \cos \theta=\frac{\sin \theta}{2} \\ \therefore \quad & \tan \theta=4 \quad \Rightarrow \theta=\tan ^{-1}(4)\end{aligned}$

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

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