A projectile is thrown at an angle $\theta$ such that it is just able to cross a vertical wall at its highest point as shown in the figure.The angle $\theta$ at which the projectile is thrown is given by

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  • A

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

  • B

    $\tan ^{-1}(\sqrt{3})$

  • C

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

  • D

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

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  • [AIPMT 2000]