WEBVTT 00:00:00.000 --> 00:00:03.000 Let's start off with this problem--let's pick a location to define 00:00:03.000 --> 00:00:06.000 a potential energy U to be equal to 0. 00:00:06.000 --> 00:00:10.000 I'm going to choose just this line here, a place where the skier will land 00:00:10.000 --> 00:00:13.000 and where all of our heights are measured from just to make things nice and easy. 00:00:13.000 --> 00:00:17.000 The first part of the problem, I want to figure out what the speed of the skier is 00:00:17.000 --> 00:00:21.000 at this point right here when he finally just leaves ramp 00:00:21.000 --> 00:00:25.000 because that is just a projectile motion problem, and we know how to do that from unit 2. 00:00:25.000 --> 00:00:28.000 I want to figure out using energy techniques. 00:00:28.000 --> 00:00:33.000 What his kinetic energy and therefore what his speed will be when he's leaving the ramp right here. 00:00:33.000 --> 00:00:38.000 Now, we know from conservation of energy that the skier's kinetic and potential energy 00:00:38.000 --> 00:00:43.000 must add up to the same value in both his starting position here and when he's just leaving the ramp. 00:00:43.000 --> 00:00:46.000 We can actually get rid one of these four variables already. 00:00:46.000 --> 00:00:51.000 At the beginning of the jump, the skier isn't moving yet, so we know his kinetic energy must be 0. 00:00:51.000 --> 00:00:56.000 Doing a little Algebra, I find that his final kinetic energy should be equal to the difference 00:00:56.000 --> 00:01:00.000 between his original potential energy and his final potential energy. 00:01:00.000 --> 00:01:03.000 And doing a little algebra and plugging in the values we were given, 00:01:03.000 --> 00:01:07.000 we find that the skier's velocity is 40 m/s when he's leaving the ramp here. 00:01:07.000 --> 00:01:09.000 Now that I know the speed of the skier at the tip of the ramp, 00:01:09.000 --> 00:01:14.000 I can now calculate using kinematics how far away from the ramp he will land. 00:01:14.000 --> 00:01:17.000 Note that in the previous part of the problem, 40 m/s was what we called 00:01:17.000 --> 00:01:19.000 the final speed, and it was for that part. 00:01:19.000 --> 00:01:22.000 It was what the skier's speed was at the end of the first part of the problem 00:01:22.000 --> 00:01:24.000 when he was just on the tip of the ramp. 00:01:24.000 --> 00:01:28.000 But now, that 40 m/s is the original speed for the second part of the problem 00:01:28.000 --> 00:01:31.000 when the skier is flying through the air. 00:01:31.000 --> 00:01:34.000 That's why I'm using this convention of calling this V0 now. 00:01:34.000 --> 00:01:37.000 The first thing I'm going to do is break this original velocity 00:01:37.000 --> 00:01:40.000 into x and y components based on this angle α. 00:01:40.000 --> 00:01:49.000 And in doing so, I find that the original Vx is 34.64 m/s, and the original Vy is 20 m/s. 00:01:49.000 --> 00:01:53.000 Now, I'm going to use this kinematic equation to solve for the time t 00:01:53.000 --> 00:02:00.000 when the skier lands on the ground and that's when his change in y would be -10 since 10 appeared 00:02:00.000 --> 00:02:06.000 and then he goes down to 0, a change of -10, and doing a little algebra, I write this quadratic equation. 00:02:06.000 --> 00:02:09.000 Now, it doesn't look like I can factor this quadratic equation, 00:02:09.000 --> 00:02:11.000 so I'm going to have to use the quadratic formula. 00:02:11.000 --> 00:02:13.000 And plugging these numbers into the quadratic equation, 00:02:13.000 --> 00:02:21.000 I get two solutions, t=-0.45 s and t=4.45 s. 00:02:21.000 --> 00:02:24.000 Now, what's the deal with this negative time one? 00:02:24.000 --> 00:02:27.000 Well, if we look back at our graph, we can actually imagine 00:02:27.000 --> 00:02:31.000 the skier going backwards in time on his trajectory, prodopdopdop. 00:02:31.000 --> 00:02:36.000 And he would actually intercept the y equal 0 line right about here. 00:02:36.000 --> 00:02:41.000 Our solution is actually still correct. It's not quite what we're looking for in this problem. 00:02:41.000 --> 00:02:46.000 Instead we're going to focus on this positive solution, t=4.45 s, and that sounds reasonable. 00:02:46.000 --> 00:02:51.000 It sounds like it might take about 4.5 s for the skier to go in the air and land back on the ground. 00:02:51.000 --> 00:02:56.000 Okay. Now, I'm going to use this equation to figure out how far in the x direction 00:02:56.000 --> 00:02:59.000 the skier travels during that time he is in the air. 00:02:59.000 --> 00:03:03.000 Now keep in mind there's no acceleration in the x direction so this term goes to 0, 00:03:03.000 --> 00:03:07.000 and we are left with just Δx=V0x*t. 00:03:07.000 --> 00:03:12.000 And plugging in the numbers we were given, we got that the final change in x. 00:03:12.000 --> 99:59:59.999 The distance d that the skier lands away from the ramp is 154.1 m. If you got this answer, great work