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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?

enter image description here

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.

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perpendicular components of the object's velocity are independent, so the one which has the smallest initial velocity perpendicular to the ground will hit the floor first. –  user27182 Sep 2 '12 at 20:31
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@Hayeder: And another way to put it is height. The one that rises the least will hit the ground first. –  Mike Dunlavey Sep 3 '12 at 1:49
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It will hit the ground quickest if you point it to -90°. I.e. straight down. –  Mark Adler Sep 3 '12 at 4:11
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4 Answers 4

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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.

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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 $t$.

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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.

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$v_{0y}$ would be greater for cannon A in your picture, so A would go higher and stay in the air longer.

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