AB is an inclined plane of inclination $30°$ with horizontal. Point O is 20m above point A. A particle is projected horizontally and it collides  perpendicular to the plane AB . Speed of the particle, at ‘O’ must be  $\left(g=10{\text{\hspace{0.17em}m/s}}^{\text{2}}\right)$

Solution:

Resolving initial velocity and  g along  x and  y direction
We get  $u_x=u\cos \text{θ,\hspace{0.17em}\hspace{0.17em}}{u}_y=u\sin {\text{θ,\hspace{0.17em}\hspace{0.17em}θ=30}}^{\circ }$
$g_x=-g\sin \text{θ}$
$g_y=g\cos \text{θ}$
From diagram
Applying equation of motion in  x-direction
$v_x=0$ ($\because$ body hits the plane normally )
$u_x=u\cos \text{θ}$

$\Rightarrow v_x=u_x+a_{x}t$

$\Rightarrow 0=u\cos 30°-g\sin 30°t$

$\Rightarrow u\sqrt{3}=gt$

$\Rightarrow t=\frac{\sqrt{3}u}{g}$……. (1)

Applying equation of motion in y direction

$\begin{array}{l}\Rightarrow x_y=u_{y}t+\frac{1}{2}{a}_y{t}^2\\ \Rightarrow 20\cos {30}^{\circ }=u\sin {30}^{\circ }t+\frac{1}{2}g\cos {30}^{\circ }{t}^2\dots \dots \text{ (2) }\\ \Rightarrow 20\sqrt{3}=ut+\frac{1}{2}g{t}^2\sqrt{3}\\ \Rightarrow 20\sqrt{3}=\frac{u^2\sqrt{3}}{g}+\frac{\sqrt{3}}{2}g\cdot \frac{3u^2}{g^2}(\because g=10\mathrm{m}/\mathrm{s})\\ \Rightarrow 20\sqrt{3}=\frac{u^2\left(2\sqrt{3}+3\sqrt{3}\right)}{2g}\\ \Rightarrow 20\sqrt{3}=\frac{u^2\left(5\sqrt{3}\right)}{2g}\\ \Rightarrow u^2=80\end{array}$

Solving the above equation we get  $u=4\sqrt{5}\text{\hspace{0.17em}m}/\text{s}$
Therefore, the correct answer is (C).