One second after the projection, a stone moves at an angle of 45⁰ with the horizon. After two seconds, it is found to move horizontally. The angle of projection is,

The correct option is D $ta{n}^{-1}\left(2\right)$
When stone travels horizontally, then it must be at the maximum height.
So, $2=\frac{usin\theta }{g}$
$\Rightarrow usin\theta =20----\left(i\right)$
Also, $tan{45}^{\circ }=\frac{usin\theta -g}{ucos\theta }$
$\Rightarrow ucos\theta =20-10\Rightarrow ucos\theta =10----\left(ii\right)$
Squaring Eqs. (i) and (ii) and adding, we get
$u=10\sqrt{5}m{s}^{-1}$ and $tan\theta =2$