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