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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
$\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$\tan ^{-1}(\sqrt{3})$
$\tan ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
$\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)$
Solution
(c)
$\frac{R / 2}{H}=\frac{\sqrt{3} H}{H}=\sqrt{3}$
or $\frac{\left(v_0^2 \sin \theta \cos \theta\right) / g}{\left(v_0^2 \sin ^2 \theta\right) / 2 g}=\sqrt{3}$
$2 \cot \theta=\sqrt{3}$
or $\tan \theta=\frac{2}{\sqrt{3}}$
or $\theta=\tan ^{-1}\left(\frac{2}{\sqrt{3}}\right)$
Similar Questions
Trajectory of particle in a projectile motion is given as $y=x-\frac{x^2}{80}$. Here, $x$ and $y$ are in metre. For this projectile motion match the following with $g=10\,m / s ^2$.
| $Column-I$ | $Column-II$ |
| $(A)$ Angle of projection | $(p)$ $20\,m$ |
| $(B)$ Angle of velocity with horizontal after $4\,s$ | $(q)$ $80\,m$ |
| $(C)$ Maximum height | $(r)$ $45^{\circ}$ |
| $(D)$ Horizontal range | $(s)$ $\tan ^{-1}\left(\frac{1}{2}\right)$ |
