A ball falls onto an inclined plane of inclination θ from a height...

Velocity of the ball just before impact is $v = \sqrt{2gh}$ vertically downwards.

Taking the x-axis along the inclined plane downwards and the y-axis perpendicular to the plane upwards.

The components of velocity just before impact are:
$v_{x} = v \sin\theta$
$v_{y} = -v \cos\theta$

Since the collision is perfectly elastic, the velocity component perpendicular to the plane is reversed, while the parallel component remains unchanged. The velocity components just after impact are:
$u_{x} = v \sin\theta$
$u_{y} = v \cos\theta$

The components of acceleration due to gravity are:
$a_{x} = g \sin\theta$
$a_{y} = -g \cos\theta$

The time of flight $T$ before the ball strikes the plane again is obtained by setting the displacement along the y-axis to zero:
$y = u_{y}T + \dfrac{1}{2}a_{y}T^2 = 0$
$v \cos\theta T - \dfrac{1}{2}g \cos\theta T^2 = 0$
$T = \dfrac{2v}{g}$

The distance $R$ along the incline is the displacement along the x-axis in time $T$:
$R = u_{x}T + \dfrac{1}{2}a_{x}T^2$
$R = (v \sin\theta)\left(\dfrac{2v}{g}\right) + \dfrac{1}{2}(g \sin\theta)\left(\dfrac{2v}{g}\right)^2$
$R = \dfrac{2v^2 \sin\theta}{g} + \dfrac{2v^2 \sin\theta}{g} = \dfrac{4v^2 \sin\theta}{g}$

Substituting $v^2 = 2gh$:
$R = \dfrac{4(2gh)\sin\theta}{g} = 8h \sin\theta$