Note the x-y coordinate system of the applet.  The origin is at the center of the MGR, +x is to the right, and +y is up.  We'll use the following symbols for brevity:

R = radius of merry-go-round (fixed at 10 m)
T = period of merry-go-round
Vo = initial throw speed of ball relative to Blue rider
q = initial throw direction of ball (0° is toward center of MGR, +90° is in +y direction)
Vr = speed of either rider relative to the ground* (Note that Vr = 2pR / T)

*For simplicity, we'll sometimes refer to the Inertial Frame as the "ground".

In order to help keep Frames of Reference straight, we'll change to blue italics when discussing the Non-Inertial Frame, that is, the frame that includes the MGR and its riders.

1.  Click on Initialize/Reset to set the parameters to their original values.  Toggle to the Inertial Frame, if necessary.  Now click Forward.  Note that even though Blue threw the ball toward the center of the MGR, the direction of the ball's path is diagonally toward the top of the screen.  That's because the ball has, in addition to an x-component of velocity, a y-component equal to Blue's velocity at the instant of the throw.  These components are designated Vx and Vy in the applet window.  Note that Vy = Vr = 2p(10 m) / (5 s) = 12.56 m/s.  Notice also that the path of the ball is a straight line, since Vx and Vy do not change.  That makes sense because there's no force acting on the ball after it leaves Blue's hand.

2.  Now Pause and Reset the applet, keeping the same values of T, Vo, and qToggle to the Non-Inertial Frame.  Try to predict what path the ball will take from this reference frame.  Remember, this is like the path seen by a rider on the MGR.  When you're ready, click Forward.  Why doesn't the MGR move?  (The riders view the rest of the world as moving while they are stationary.)  Notice that the ball starts out heading directly for the center of the MGR but then veers away.   It seems as if there is a force accelerating the ball away from the MGR. Yet, we saw for the Inertial Frame that there was no force on the ball.  From an Inertial point of view, we can explain the apparent force on the ball seen by the riders as being due to their own rotation.

3.  Pause the applet, increase Vo  to 50 m/s and Reset.  Toggle to the Inertial Frame.  What do you expect the path of the ball to be this time?  (The Vy will be the same as before since Blue's speed has not changed.  However, the Vx is 5 times greater than before.)  Start the applet.  Were you correct?   The ball barely misses the center of the MGR.  (Do you see why the ball cannot pass directly through the center as long as q = 0°?)  Now what would this look like from the Non-Inertial Frame?  Let's predict first.  We know the ball will miss the center of the MGR.  That result has to be the same in both frames.  In the Non-Inertial Frame, however, the ball's initial velocity vector is directly toward the center.  Therefore, if the ball is to miss, it will have to veer away.  Try it and see.

4.  What happens when you swing a ball in a circle at the end of a tether and suddenly let go?  The ball travels tangent to the circular path that the ball had.   The applet can demonstrate this.  Set Vo to 0.  This means that Blue simply lets go of the ball.  Reset the applet, toggle to the Inertial Frame, and click Forward.  As the motion progresses, try to imagine what the path of the ball would look like from the Non-Inertial Frame.  Then Pause, Reset, toggle to the Non-Inertial Frame, and see if you're right.

5.  Pause the applet and toggle to the Inertial Frame.  What would Vo and q have to be so that the ball wouldn't move from its initial position at (x,y) = (R,0)?  That is, we want the velocity of the ball relative to the ground to be 0. Enter your predicted values, Reset, and click Forward.  Were you right?  Hopefully, you realized that the velocity of the ball relative to Blue would have to be the opposite of Blue's initial velocity relative to the ground.  The latter is 12.56 m/s in the +y direction.   Therefore, the ball's velocity relative to Blue must be 12.56 m/s in the -y direction (q = 90°).  Now predict what the path would look like from the Non-Inertial Frame and try it.

If you're ready, go on to some quantitative problems ...