Answer
The ball lands on the ramp.
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.
