How far from the bottom of the chute does the ball land? your answer for the distance the ball travels from the end of the chute should contain r?

To get the distance, find the unchanging horizontal component of velocity and the time of flight and multiply them.

First find the horizontal component of velocity, Vh by using the equation for centripetal acceleration: v² / r = a Vh² / R = 2g Vh = 2gR 
Therefore the horizontal component of velocity is: 
►Vh = √( 2gR ) 

Now to find time elapsed. In the vertical dimension, the ball started its flight with no initial velocity. We determine how far it travels vertically by the equation: 
d = ½at² 
And substitute our values for d and a and solve for time: 
2R = ½gt² t² = 4R / g t = √( 4R / g ) t = 2√( R / g ) 
►∆t = 2√( R / g ) 
Now we multiply our two critical values together to get the distance: 
d = Vh•∆t 
d = [ √( 2gR ) ]•[ 2√( R / g ) ] 
d = 2√( 2 R² ) 
d = 2√2 R

Ithink they do have the same properties just in less quantities i think