We will work this out using the English System of measurement. If the football field was 100 meters long instead of 100 yards (It would only be 9 yards longer), the metric system would probably catch on in the U.S. a lot faster. Alas, that won't be happening any time soon.
So we let d = 300 feet, g = 32 feet per second, the angle is the optimum throwing angle of 45 degrees, the initial height is 6 feet, which I figure is about where the ball would leave the quarterback's hand. The velocity is what it would take to throw it that far.
This is not the way to go with this. For one thing, solving it is really hard. Trust me. I did it and its rough. You could assign it for a massive amount of extra credit, but otherwise it isn't worth it. Secondly, I didn't take into account that we are measuring to where it hits the ground which is not the same elevation that it took off anyway.
Let's just take a starting height of zero. That will make things easier and it turns out demonstrates an nice application of a trig identity.
So here we go. With a starting height of zero, the above equation becomes:
d = (vcos(θ)/g)ᆞ( vsin(θ) + vsin(θ))
d = (vcos(θ)/g)ᆞ(2vsin(θ))
d = (v2/g)ᆞ(2sin(θ)cos(θ))
But, there is a double angle identity that states: sin(2θ) = 2sin(θ)cos(θ). So our formula becomes:
d = (v2/g)ᆞ(sin(2θ))
This then becomes the formula that is commonly found for range of a projectile with a starting height of zero.
Nice application of an identity.
I feel that is quite enough excitement for one day. Next time we'll answer the question of whether Roman likely threw the ball 100 yards.
