﻿[Script Info] Title: [Events] Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text Dialogue: 0,0:00:00.00,0:00:03.00,Default,,0000,0000,0000,,Well the answer is about 5.5° and let's see how we calculate that. Dialogue: 0,0:00:03.00,0:00:07.00,Default,,0000,0000,0000,,Well first, we need to calculate the range of the cannon. How do we do that? Dialogue: 0,0:00:07.00,0:00:09.00,Default,,0000,0000,0000,,Let's look at our initial velocity vector. Dialogue: 0,0:00:09.00,0:00:14.00,Default,,0000,0000,0000,,The cannon fires at the speed of 56 m/s, an angle of 45° because that optimizes the range. Dialogue: 0,0:00:14.00,0:00:18.00,Default,,0000,0000,0000,,Now we want to break this into components--V₀x and V₀y. Dialogue: 0,0:00:18.00,0:00:24.00,Default,,0000,0000,0000,,We can do some trigonometry to show that the x and y components are both equal to 39.6 m/s. Dialogue: 0,0:00:24.00,0:00:26.00,Default,,0000,0000,0000,,If I'm going to know how far this cannonball is going to go, I should Dialogue: 0,0:00:26.00,0:00:29.00,Default,,0000,0000,0000,,probably know how long it's going to spend in the air. Dialogue: 0,0:00:29.00,0:00:31.00,Default,,0000,0000,0000,,Let's use the y component velocity to do that. Dialogue: 0,0:00:31.00,0:00:36.00,Default,,0000,0000,0000,,Looking at the trajectory of this ball, I want to first find out how long it takes to get to the very top Dialogue: 0,0:00:36.00,0:00:39.00,Default,,0000,0000,0000,,that's when the vertical component of velocity would be equal to zero. Dialogue: 0,0:00:39.00,0:00:43.00,Default,,0000,0000,0000,,Using this equation and plugging in what we know, and remember in this case acceleration Dialogue: 0,0:00:43.00,0:00:48.00,Default,,0000,0000,0000,,is downward so negative, we find it takes 3.96 seconds to get to the top of the trajectory, Dialogue: 0,0:00:48.00,0:00:55.00,Default,,0000,0000,0000,,which means it would take another 3.96 to get down which results in a total time of 7.92 seconds. Dialogue: 0,0:00:55.00,0:00:58.00,Default,,0000,0000,0000,,Now we known the x direction, speed isn't going to be changing. Dialogue: 0,0:00:58.00,0:01:00.00,Default,,0000,0000,0000,,There is no acceleration in the x direction. Dialogue: 0,0:01:00.00,0:01:04.00,Default,,0000,0000,0000,,So if I look at this equation, this term goes to 0 and we're left with this. Dialogue: 0,0:01:04.00,0:01:08.00,Default,,0000,0000,0000,,So what's the range of the cannon, how far can it reach Δx or it's just Dialogue: 0,0:01:08.00,0:01:14.00,Default,,0000,0000,0000,,the x component of velocity 39.6 times the time, which gives us 314 m. Dialogue: 0,0:01:14.00,0:01:17.00,Default,,0000,0000,0000,,If you got this far, good work. We're almost towards the end. Dialogue: 0,0:01:17.00,0:01:20.00,Default,,0000,0000,0000,,Now we just need to convert this somehow into an angular size. Dialogue: 0,0:01:20.00,0:01:24.00,Default,,0000,0000,0000,,Well looking at this angle β, we know this flagpole has a height of 30 m Dialogue: 0,0:01:24.00,0:01:28.00,Default,,0000,0000,0000,,and in fact, it looks like we know the opposite and the adjacent sides. Dialogue: 0,0:01:28.00,0:01:31.00,Default,,0000,0000,0000,,So we can write the tangent of β is 30/314. Dialogue: 0,0:01:31.00,0:01:36.00,Default,,0000,0000,0000,,If I ask Google what the arctan of 30/314, in degrees because remember that's how we go Dialogue: 0,0:01:36.00,9:59:59.99,Default,,0000,0000,0000,,from ratios to angles, it gives me a value of 5.5°--good work!