Which angle causes an object to land quickest? Say a canon is where the circle is, and it shoots two different canonballs at different angles, but at the same speed, which angle would make the cannonball hit the ground first?

Intuitively I'd think they'd hit at the same time, however, I remember a formula describing the second coordinate $y$: $y = - 1/2 \cdot g \cdot t^2 + v_{0y} \cdot t$
Where $t$ is the time, and $v_{0y}$ is the starting velocity. Thus a high $v_{0y}$, would lead to a greater total time $t$, thus it would hit A first. This seems to be missing something though.
 A: For a fixed muzzle velocity the time the cannonball stays in the air depends on the vertical component of the velocity, so trajectory A would stay in the air longest. The trajectory with the longest duration is firing directly upwards.
If the angle to the ground is $\theta$, then the vertical velocity is $v_0sin\theta$, and the time in the air (neglecting air resistance) is $2v_0sin\theta/g$, where $g$ is the acceleration due to gravity.
A: Set the equation $y = -0.5gt^2 + V_{y_0} t$ equal to zero. You get two solutions one at time $t = 0$, and the second one is when the object hits the ground again at $t = 2V_{y_0}/g$. Of course, $V_{y_0} = V_0 \sin(\theta)$. Thus you have: $t (\text{travel}) = 2 V_0 \sin(\theta)/g$. The closer the angle to 90 degree the longer the flight time.
A: You are ignoring here the resistance of air, right? I am not going to give you the solution which you can find in any physics text book, but rather tell you how to derive it yourself. 
Try expressing initial speed $v_0$ as a function of your angle $\alpha$ (hint: sines and cosines), then solve for $t$ in your formula. From there the solution should be fairly obvious, but you can always verify by finding the angle $\alpha$ which corresponds to smallest $t$.
A: $v_{0y}$ would be greater for cannon A in your picture, so A would go higher and stay in the air longer.
