Hey, I totally missed ths post! My Master's degree could have been handy here
j4ck is totally correct on his explanation without knowing some information. It is unclear what happens before the ball leves the table. It seems that the ball is released from the inside of a curved track exactly at the middle of the track. So the initial PE (potential energy) would be MgR + MgH. This is the total energy of the system.
for a) I believe you mean the velocity of the ball when it leaves the table. This can be solved from v = sqrt (2gR).
for b) remember Galileo's superposition principle that you treat the horizontal and vertical movements independent of each other. Basically you know initial velocity = 0, acceleration in vertical direction is g, and the height it falls is H. You know three things so you can solve for both the time and final velocity. Use the equation H = vt + (1/2) g t^2. Where v = intial velocity. Therfeore t =sqrt (2 H /g)
for c) remeber that the horizontal velocity never changes so v = x / t. So the distance x it goes is just given by x = v t, where v is from part a) and t is from part b)
for d) the total amount of energy when the ball hits the floor is just the initial PE with respect to the floor. PE = MgR + MgH = Mg (R + H)
I am guessing at the initial positionof the ball, but I have seen similar problems to the one you needed help with. Good luck in your class!