[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,,Let's start off with this problem--let's pick a location to define
Dialogue: 0,0:00:03.00,0:00:06.00,Default,,0000,0000,0000,,a potential energy U to be equal to 0.
Dialogue: 0,0:00:06.00,0:00:10.00,Default,,0000,0000,0000,,I'm going to choose just this line here, a place where the skier will land
Dialogue: 0,0:00:10.00,0:00:13.00,Default,,0000,0000,0000,,and where all of our heights are measured from just to make things nice and easy.
Dialogue: 0,0:00:13.00,0:00:17.00,Default,,0000,0000,0000,,The first part of the problem, I want to figure out what the speed of the skier is
Dialogue: 0,0:00:17.00,0:00:21.00,Default,,0000,0000,0000,,at this point right here when he finally just leaves ramp
Dialogue: 0,0:00:21.00,0:00:25.00,Default,,0000,0000,0000,,because that is just a projectile motion problem, and we know how to do that from unit 2.
Dialogue: 0,0:00:25.00,0:00:28.00,Default,,0000,0000,0000,,I want to figure out using energy techniques.
Dialogue: 0,0:00:28.00,0:00:33.00,Default,,0000,0000,0000,,What his kinetic energy and therefore what his speed will be when he's leaving the ramp right here.
Dialogue: 0,0:00:33.00,0:00:38.00,Default,,0000,0000,0000,,Now, we know from conservation of energy that the skier's kinetic and potential energy
Dialogue: 0,0:00:38.00,0:00:43.00,Default,,0000,0000,0000,,must add up to the same value in both his starting position here and when he's just leaving the ramp.
Dialogue: 0,0:00:43.00,0:00:46.00,Default,,0000,0000,0000,,We can actually get rid one of these four variables already.
Dialogue: 0,0:00:46.00,0:00:51.00,Default,,0000,0000,0000,,At the beginning of the jump, the skier isn't moving yet, so we know his kinetic energy must be 0.
Dialogue: 0,0:00:51.00,0:00:56.00,Default,,0000,0000,0000,,Doing a little Algebra, I find that his final kinetic energy should be equal to the difference
Dialogue: 0,0:00:56.00,0:01:00.00,Default,,0000,0000,0000,,between his original potential energy and his final potential energy.
Dialogue: 0,0:01:00.00,0:01:03.00,Default,,0000,0000,0000,,And doing a little algebra and plugging in the values we were given,
Dialogue: 0,0:01:03.00,0:01:07.00,Default,,0000,0000,0000,,we find that the skier's velocity is 40 m/s when he's leaving the ramp here.
Dialogue: 0,0:01:07.00,0:01:09.00,Default,,0000,0000,0000,,Now that I know the speed of the skier at the tip of the ramp,
Dialogue: 0,0:01:09.00,0:01:14.00,Default,,0000,0000,0000,,I can now calculate using kinematics how far away from the ramp he will land.
Dialogue: 0,0:01:14.00,0:01:17.00,Default,,0000,0000,0000,,Note that in the previous part of the problem, 40 m/s was what we called
Dialogue: 0,0:01:17.00,0:01:19.00,Default,,0000,0000,0000,,the final speed, and it was for that part.
Dialogue: 0,0:01:19.00,0:01:22.00,Default,,0000,0000,0000,,It was what the skier's speed was at the end of the first part of the problem
Dialogue: 0,0:01:22.00,0:01:24.00,Default,,0000,0000,0000,,when he was just on the tip of the ramp.
Dialogue: 0,0:01:24.00,0:01:28.00,Default,,0000,0000,0000,,But now, that 40 m/s is the original speed for the second part of the problem
Dialogue: 0,0:01:28.00,0:01:31.00,Default,,0000,0000,0000,,when the skier is flying through the air.
Dialogue: 0,0:01:31.00,0:01:34.00,Default,,0000,0000,0000,,That's why I'm using this convention of calling this V0 now.
Dialogue: 0,0:01:34.00,0:01:37.00,Default,,0000,0000,0000,,The first thing I'm going to do is break this original velocity
Dialogue: 0,0:01:37.00,0:01:40.00,Default,,0000,0000,0000,,into x and y components based on this angle α.
Dialogue: 0,0:01:40.00,0:01:49.00,Default,,0000,0000,0000,,And in doing so, I find that the original Vx is 34.64 m/s, and the original Vy is 20 m/s.
Dialogue: 0,0:01:49.00,0:01:53.00,Default,,0000,0000,0000,,Now, I'm going to use this kinematic equation to solve for the time t
Dialogue: 0,0:01:53.00,0:02:00.00,Default,,0000,0000,0000,,when the skier lands on the ground and that's when his change in y would be -10 since 10 appeared
Dialogue: 0,0:02:00.00,0:02:06.00,Default,,0000,0000,0000,,and then he goes down to 0, a change of -10, and doing a little algebra, I write this quadratic equation.
Dialogue: 0,0:02:06.00,0:02:09.00,Default,,0000,0000,0000,,Now, it doesn't look like I can factor this quadratic equation,
Dialogue: 0,0:02:09.00,0:02:11.00,Default,,0000,0000,0000,,so I'm going to have to use the quadratic formula.
Dialogue: 0,0:02:11.00,0:02:13.00,Default,,0000,0000,0000,,And plugging these numbers into the quadratic equation,
Dialogue: 0,0:02:13.00,0:02:21.00,Default,,0000,0000,0000,,I get two solutions, t=-0.45 s and t=4.45 s.
Dialogue: 0,0:02:21.00,0:02:24.00,Default,,0000,0000,0000,,Now, what's the deal with this negative time one?
Dialogue: 0,0:02:24.00,0:02:27.00,Default,,0000,0000,0000,,Well, if we look back at our graph, we can actually imagine
Dialogue: 0,0:02:27.00,0:02:31.00,Default,,0000,0000,0000,,the skier going backwards in time on his trajectory, prodopdopdop.
Dialogue: 0,0:02:31.00,0:02:36.00,Default,,0000,0000,0000,,And he would actually intercept the y equal 0 line right about here.
Dialogue: 0,0:02:36.00,0:02:41.00,Default,,0000,0000,0000,,Our solution is actually still correct. It's not quite what we're looking for in this problem.
Dialogue: 0,0:02:41.00,0:02:46.00,Default,,0000,0000,0000,,Instead we're going to focus on this positive solution, t=4.45 s, and that sounds reasonable.
Dialogue: 0,0:02:46.00,0:02:51.00,Default,,0000,0000,0000,,It sounds like it might take about 4.5 s for the skier to go in the air and land back on the ground.
Dialogue: 0,0:02:51.00,0:02:56.00,Default,,0000,0000,0000,,Okay. Now, I'm going to use this equation to figure out how far in the x direction
Dialogue: 0,0:02:56.00,0:02:59.00,Default,,0000,0000,0000,,the skier travels during that time he is in the air.
Dialogue: 0,0:02:59.00,0:03:03.00,Default,,0000,0000,0000,,Now keep in mind there's no acceleration in the x direction so this term goes to 0,
Dialogue: 0,0:03:03.00,0:03:07.00,Default,,0000,0000,0000,,and we are left with just Δx=V0x*t.
Dialogue: 0,0:03:07.00,0:03:12.00,Default,,0000,0000,0000,,And plugging in the numbers we were given, we got that the final change in x.
Dialogue: 0,0:03:12.00,9:59:59.99,Default,,0000,0000,0000,,The distance d that the skier lands away from the ramp is 154.1 m. If you got this answer, great work