A physics question I made up

I like to lay down on my floor and throw a ball at my ceiling. The distance from my release point to the ceiling is 76 inches. If I toss it up at the ceiling such that it barely touches it, how fast did I throw the ball?

I've been out of physics (although recently took diff eq.) for a while and this question is driving me crazy! How in the world do I get this? Explanations are crucial, thank you.

Ack, inches! ;-) I'll do it in metres and you can convert back to imperial units at the end if you like!

76 inches = 1.93 m

Using -9.8 m/s/s for the acceleration due to gravity, and u for the initial velocity you get:

d = ut + 1/2 at^2

1.93 = ut - 4.9 t^2 (the gravity is negative because it is in the opposite direction to the initial velocity)

Still got two unknowns, so we need another equation. Remember that the velocity will be 0 at the top of the ball's motion:

v = u + at

0 = u - 9.8 t

That means u = 9.8t, so we can sub that in the other equation and get it all in terms of t:

1.93 = 9.8t^2 - 4.9t^2

1.93 = 4.9t^2

t^2 = 1.93/4.9

Gives t = 0.63 s

Now we know u = 9.8t = 9.8(0.63) = 6.17 m/s

(76 x 0.0254) = 1.9304 metres.
Launch velocity = sqrt. (2gh), = 6.1511m/sec.
(6.1511 x 3.2808) = 20.18ft/ sec.

I don;t know but can you please answer my question ?! I need it by tomorrow if you can help please ?!