Find initial velocity of projectile - 10 points best answer

I need help with this question. I don't really understand where to start. The question is:
The quarterback of a football team releases a pass at a height of 7 feet above the
playing field, and the football is caught by a receiver 30 yards directly downfield at a height of
4 feet. The pass is released at an angle of 30 degrees with the horizontal. At what speed was
the football released?
Any help is appreciated

Let the point at which the ball was released = (0, 7)
Let the point at which the ball was caught = (90, 4)
Let g = the acceleration due to gravity = 32.2 ft/s²
Let V = the initial velocity.
Let Vy = the initial velocity in the y direction = (V)sin(30°) = .5V
Let Vx = the velocity in the x direction = (V)cos(30°) ={(√3)/2}V
The height, y(t), is:
y(t) = 7 + (0.5V)t - (1/2)(32.2)t²
The horizontal displacement is:
x(t) = {(√3)/2}V(t)
The time of flight is:
t = 60(√3)/V
Substitute this for t and make the height be 4:
4 = 7 + (0.5V)(60√3)/V - 16.1{(60√3)/V}²
-54.96 = -173880/V²
V = √{-173880/-54.96} = 56.24 ft/s

The time of flight is not needed

V² = -g*x²/[2cos²Θ(y - xtanΘ - yi)] = 3163.67 ft²/sec²

V = 56.25 m/s