1 00:00:00,000 --> 00:00:03,000 Well the answer is about 5.5° and let's see how we calculate that. 2 00:00:03,000 --> 00:00:07,000 Well first, we need to calculate the range of the cannon. How do we do that? 3 00:00:07,000 --> 00:00:09,000 Let's look at our initial velocity vector. 4 00:00:09,000 --> 00:00:14,000 The cannon fires at the speed of 56 m/s, an angle of 45° because that optimizes the range. 5 00:00:14,000 --> 00:00:18,000 Now we want to break this into components--V₀x and V₀y. 6 00:00:18,000 --> 00:00:24,000 We can do some trigonometry to show that the x and y components are both equal to 39.6 m/s. 7 00:00:24,000 --> 00:00:26,000 If I'm going to know how far this cannonball is going to go, I should 8 00:00:26,000 --> 00:00:29,000 probably know how long it's going to spend in the air. 9 00:00:29,000 --> 00:00:31,000 Let's use the y component velocity to do that. 10 00:00:31,000 --> 00:00:36,000 Looking at the trajectory of this ball, I want to first find out how long it takes to get to the very top 11 00:00:36,000 --> 00:00:39,000 that's when the vertical component of velocity would be equal to zero. 12 00:00:39,000 --> 00:00:43,000 Using this equation and plugging in what we know, and remember in this case acceleration 13 00:00:43,000 --> 00:00:48,000 is downward so negative, we find it takes 3.96 seconds to get to the top of the trajectory, 14 00:00:48,000 --> 00:00:55,000 which means it would take another 3.96 to get down which results in a total time of 7.92 seconds. 15 00:00:55,000 --> 00:00:58,000 Now we known the x direction, speed isn't going to be changing. 16 00:00:58,000 --> 00:01:00,000 There is no acceleration in the x direction. 17 00:01:00,000 --> 00:01:04,000 So if I look at this equation, this term goes to 0 and we're left with this. 18 00:01:04,000 --> 00:01:08,000 So what's the range of the cannon, how far can it reach Δx or it's just 19 00:01:08,000 --> 00:01:14,000 the x component of velocity 39.6 times the time, which gives us 314 m. 20 00:01:14,000 --> 00:01:17,000 If you got this far, good work. We're almost towards the end. 21 00:01:17,000 --> 00:01:20,000 Now we just need to convert this somehow into an angular size. 22 00:01:20,000 --> 00:01:24,000 Well looking at this angle β, we know this flagpole has a height of 30 m 23 00:01:24,000 --> 00:01:28,000 and in fact, it looks like we know the opposite and the adjacent sides. 24 00:01:28,000 --> 00:01:31,000 So we can write the tangent of β is 30/314. 25 00:01:31,000 --> 00:01:36,000 If I ask Google what the arctan of 30/314, in degrees because remember that's how we go 26 00:01:36,000 --> 99:59:59,999 from ratios to angles, it gives me a value of 5.5°--good work!