Answer
The ball does not land on the ramp because the value of vertical height is less then the height of plateau.
Work Step by Step
We can find the time to travel a horizontal distance of $6.00~m$:
$x = v_x~t$
$t = \frac{x}{v_x}$
$t = \frac{6.00~m}{(10.0~m/s)~cos~50.0^{\circ}}$
$t = 0.9334~s$
We can find the vertical height at this time:
$y = v_{y0}~t+\frac{1}{2}a_y~t^2$
$y = (10.0~m/s)(sin~50.0^{\circ})(0.9334~s)+\frac{1}{2}(-9.8~m/s^2)~(0.9334~s)^2$
$y = 2.88~m$
Since the ball's height is less than the height of the plateau, the ball lands on the ramp.
