A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle with speed $v_o$ and rebounds elastically in the figure. Find the distance along the plane where it will hit second time.

Formula used:
$T=\frac{2u_y}{g_{eff}}$

$s=ut+\frac{1}{2}a{t}^2$


When $x$ and $y$ axis are selected as shown in the diagram, the motion of projectile $A$ is:

$y=0$
$u_y=v_0\cos \theta$
$a_y=-\theta$
$t=T$

From Newton’s second equation of motion
$y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$0=v_0\cos \theta T+\frac{1}{2}(-g\cos \theta )T^2$
$T=\frac{2v_0}{g}$

Range along $x-$axis:
$x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
$u_x=v_o\sin \theta$
$a_x=g\sin \theta$
$x=L$
$t=T$

Putting values,we get
$L=v_o\sin \theta \left(\frac{2v_0}{g}\right)+\frac{1}{2}g\sin \theta {\left(\frac{2v_0}{g}\right)}^2$

$L=\frac{4v_o^2\sin \theta }{g}$