#### Projectile motion when projected horizontally with a speed of 20 m/s, Its angle made with the direction perpendicular to the ground  after  $t=2\sqrt{5} \ s$ is ?Option: 1 ${\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt3}\right)$Option: 2 $\ \tan^{-1}\sqrt5$Option: 3 $\tan^{-1}\sqrt3$Option: 4 $\tan^{-1}\left(\frac{1}{\sqrt5}\right)$

Answer: (4)  $\tan^{-1}\left(\frac{1}{\sqrt5}\right)$

Explanation:

$tan\theta=\frac{v_x}{v_y}$

Let,

$v_x=u$   and      $v_y=gt$

$tan\theta\ =\frac{u}{gt}$

Or, $\frac{20}{10\times2\sqrt5}$

Or,   $tan\theta=\frac{1}{\sqrt5}$

Or,   $\theta\ =\ \tan^{-1}\left(\frac{1}{\sqrt5}\right)$
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