0:00:00.000,0:00:03.000
Well the answer is about 5.5° and let's see how we calculate that.
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Well first, we need to calculate the range of the cannon. How do we do that?
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Let's look at our initial velocity vector.
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The cannon fires at the speed of 56 m/s, an angle of 45° because that optimizes the range.
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Now we want to break this into components--V₀x and V₀y.
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We can do some trigonometry to show that the x and y components are both equal to 39.6 m/s.
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If I'm going to know how far this cannonball is going to go, I should
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probably know how long it's going to spend in the air.
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Let's use the y component velocity to do that.
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Looking at the trajectory of this ball, I want to first find out how long it takes to get to the very top
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that's when the vertical component of velocity would be equal to zero.
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Using this equation and plugging in what we know, and remember in this case acceleration
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is downward so negative, we find it takes 3.96 seconds to get to the top of the trajectory,
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which means it would take another 3.96 to get down which results in a total time of 7.92 seconds.
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Now we known the x direction, speed isn't going to be changing.
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There is no acceleration in the x direction.
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So if I look at this equation, this term goes to 0 and we're left with this.
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So what's the range of the cannon, how far can it reach Δx or it's just
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the x component of velocity 39.6 times the time, which gives us 314 m.
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If you got this far, good work. We're almost towards the end.
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Now we just need to convert this somehow into an angular size.
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Well looking at this angle β, we know this flagpole has a height of 30 m
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and in fact, it looks like we know the opposite and the adjacent sides.
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So we can write the tangent of β is 30/314.
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If I ask Google what the arctan of 30/314, in degrees because remember that's how we go
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from ratios to angles, it gives me a value of 5.5°--good work!