How much force would it take to throw a 400g American football over the house drawn?

Oh wait...that's clearly Australian football as the house is on the bottom half of Earth and hanging upside down.

So we have the following constraints: it must travel X = SUM(x) = 6.1 + 7.4 + 12.4 meters in range, reach max height H at X/2, and H must be greater than 7.4 m.

As the ball is assumed caught at the same height it's tossed, the flight time T = Tu + Td = 2sqrt(2H/g) = 2*sqrt(2*7.4/9.8) = ? seconds. Then X = Ux T; so that Ux = X/T is the horizontal component of the toss.

To reach H, we must have H = Uy^2/2g where Uy is the vertical component and U^2 = Ux^2 + Uy^2 gives us the toss speed to clear the house and make that range X. So U = sqrt((X/T)^2 + 2gH) = sqrt(((6.1 + 7.4 + 12.4)/2*sqrt(2*7.4/9.8)) + 2*9.8*7.4) = 12.7 m/s is the required toss speed.

But here's the rub. As U = At = (F/M)t so that F = MU/t we need to know t, the time spent in actually throwing the ball at the speed U. As you can see, it'll take more force F to toss anything at U speed when the time spent throwing is short versus longer. BTW in case you don't see it F = MU/t is a variation of impulse.

How long far does the ball travel while that force is applied?