At what velocity does the cannonball leaves the cannon?

[Conquest Ace]

Yeah I cannot figure this out...

Homework Statement

.A cannon fires a cannonball 500.0m downrange when set at a 45.0o angle. At what velocity does the cannonball leaves the cannon?

Given Angle : 45
Distance: 500m
Vi : ?

Homework Equations

dy = viy(t)+ 4.9(t)^2

The Attempt at a Solution

Only thing I figured out is that vix = viy

and that t = 500/vicos45

 

[SHISHKABOB]

so if you know how long it was in the air, how long did it take the cannonball to get to its max height

if you know how long it took to get to the max height, then you can get the initial velocity in the y direction from that

 

[Conquest Ace]

so if you know how long it was in the air, how long did it take the cannonball to get to its max height

if you know how long it took to get to the max height, then you can get the initial velocity in the y direction from that

I think time is irreleavent here...