A particle at a height $h$ from the ground is projected with an angle ${30}^{\circ }$ from the horizontal, it strikes the ground making angle ${45}^{\circ }$ with horizontal. It is again projected from the same point with the same speed but with an angle of ${60}^{\circ }$ with horizontal. Find the angle it makes with the horizontal when it strikes the ground:

The correct option is B ${\tan }^{-1}\left(\sqrt{5}\right)$
Draw the diagram as per the given information.
diagram

Find the velocity of the particle along the y direction.
As we know,
$v_y^2=u_y^2+2gh\dots \left(i\right)$
From the diagram,
$v_y=\frac{u}{2}$
By putting the value of $v_y$ in equation $\left(i\right)$ we get,
$v_y^2={\left(\frac{u}{2}\right)}^2+2gh{v}_y=\sqrt{\frac{u^2}{4}+2gh}\dots \left(ii\right)$

Find the value of height $\left(h\right)$ from the ground.
From the given diagram we have,
therfore,
$\tan \theta =\frac{v_y}{v_x}\tan {45}^{\circ }=\frac{\sqrt{\frac{u^2}{4}+2gh}}{u\frac{\sqrt{3}}{2}}\left(iii\right)$
By putting the value of $\tan {45}^{\circ }=1$ in equation $\left(iii\right)$ we get,
$1=\frac{\sqrt{\frac{u^2}{4}+2gh}}{u\frac{\sqrt{3}}{2}}u\frac{\sqrt{3}}{2}=\sqrt{\frac{u^2}{4}+2gh}$
Squaring both sides,
$\frac{3u^2}{4}=\frac{u^2}{4}+2gh\frac{2u^2}{4}=2ghh=\frac{u^2}{4g}\dots \left(iv\right)$

Find the velocity of the particle along the $y$ direction.
From the given diagram,
$u_x=v_x=\frac{u}{2}{u}_y=\frac{\sqrt{3}u}{2}$
hence, from $3^{rd}$ equation of motion,
$v_y^2=u_y^2+2gh\dots \left(v\right)$
therefore, by putting the values we have ,
$v_y^2={\left(\frac{\sqrt{3}u}{2}\right)}^2+2g\left(\frac{u^2}{4g}\right)v_y=\frac{\sqrt{5}u}{2}$

Find the value of $\theta$
Hence,
$\tan \theta =\frac{v_y}{v_x}=\frac{\left(\frac{\sqrt{5}u}{2}\right)}{\frac{u}{2}}\tan \theta =\sqrt{5}\theta ={\tan }^{-1}\sqrt{5}$