Why is the sign negative in the kinematic equation Δy=VoyT-(1/2)gt²?

[Mindstein]

Homework Statement

A placekicker must kick a football from a point 36.0 m (about 40 yards) from the goal, and half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0 m/s at an angle of 53.0º to the horizontal.
(a) By how much does the ball clear or fall short of clearing the crossbar?
(b) Does the ball approach the crossbar while still rising or while falling?

Homework Equations

1.)Vox= Vo*CosΘ
2.)Voy= Vo*SinΘ
3.)Δx=x=VoxT where T (caps) is equal to time and not period.(its caps to discern from the sub 'ox')
4.)Δy= VoyT - (1/2)gt²

The Attempt at a Solution

Vxi = 20Cos(53)= 12.07 m/s
Vyi = 20Sin(53)= 15.97 m/s

36.0 m = (12.03 m/s)t → t = 3.00 seconds

Δy= (15.97 m/s)(3.00 seconds) - (1/2)(9.8 m/s²)(9.00 seconds²)
Δy = 3.9 meters
Clears by .9 meters...approximately
(the answer is .889 m while descending)

So, I obviously figured out how to do the problem, but I want to know why Relevant Equation 4.) requires me to use a negative sign instead of a positive one.

May someone shed some light. I am obviously missing something very similar. I tend to complicate things too much.

 

[LowlyPion]

The direction of velocity and acceleration matters.

You choose V_y as positive y, then g is going to be negative.

It starts out with V_y and all the time gravity is working to reverse it and return it to earth.

V_y gets smaller until it is 0 at max height, then it becomes negative on its downward plunge.

 

[Mindstein]

The direction of velocity and acceleration matters.

You choose V_y as positive y, then g is going to be negative.

It starts out with V_y and all the time gravity is working to reverse it and return it to earth.

V_y gets smaller until it is 0 at max height, then it becomes negative on its downward plunge.

I can not thank you enough! Thank you very much!