Kinematics of 2-D Ball Rolling Off Table

[jedjj]

2-D kinematics[solved]

Homework Statement

A ball rolls horizontally off the edge of a tabletop that is 1.90 m high. It strikes the floor at a point 1.57 m horizontally away from the table edge. (Neglect air resistance.)

(a) How long was the ball in the air?

(b) What was its speed at the instant it left the table?


2. The attempt at a solution

I have been trying to get the answer to part a for a little while now. I have been trying to use the quadratic equation to find time knowing that $$0=(-\Delta y)+V_{0y}*t+(g)t^2$$
I have not used quadratic equation in about 4 years. Am I setting this problem up correctly when I do

$$t=\frac{V_{0y}+\sqrt{V_{0y}^2-(4)(-\Delta y)(g)}}{2*(g)}$$
which should give me
$$t=\frac{{0}+\sqrt{0^2-(4)(-1.9)(g)}}{2*(g)}$$
$$t=\frac{\sqrt{-(4)(-1.9)(g)}}{2*(g)}$$

which comes to be
$$t=\frac{8.63}{19.6}$$
$$t=0.4403$$
I am not coming up with the correct answer according to what it is online[edit:still]. What am I doing wrong? Thanks for all the help.

[edit] I just went ahead and used kinematic equations and found the answer, but I'm still confused why the quadratic equation didn't work.

 

[jedjj]

To start I just realized that quadratic equation is not set up properly. I will edit the equation and set it up in the way that I believe is proper.

 

[jedjj]

Am I giving too much information, or not enough. or is no one seeing a mistake I am making?