Projectile - Speed and angle for football just clearing goalposts

[ken.drea]

Question: A kicker on a football team attempts to kick a field goal from 35 yards. The attempt just clears the goal post that is 10 feet high. The football lands 8 feet behind the plane of the goal. What speed and angle was the football initially kicked? (Use acceleration of gravity as 32 ft/s^2).

Variables given&needed
a= 32ft/s^2
Vi = ?
X= 113 ft (35yrds = 105 ft, then +8ft)
Vyf= 0

At first, I tried plugging them into this equation: Vyf^2 = Vyi^2 + 2ay
I used 10 feet as the y value. I got 25.298 as my Vyi. Then, I solved for the angle by using sine. I got an error. Then I realized that 10 feet is not the full y value, It's only the height of the goal thing.
Now I'm stuck. I was going to find time, but I'm missing a lot of variables. I was going to use it into the formula "x=Vt" to find the velocity.

Note:
a = acceleration
Vi = velocity initial
Vf = velocity final
X = horizontal distance
Y = vertical distance
t =time

 

[Simon Bridge]

Welcome to PF;
You have not used everything you know yet...
You know three points on the ball's trajectory, and you have two suvat equations for the two components of the ball's motion wrt time.

HInt: Try using the equation for x(t) to eliminate t from the equation for y(t)...